# Kinetic Theory of Gases

## Concept Of Physics

### H C Verma

1   1- Calculate the volume of 1 mole of an ideal gas at STP.

##### Solution :

Volume of 1 mole of gas $\\$PV = nRT $\Rightarrow$ $V$ = $\frac{RT}{P}$ = $\frac{0.082\times273}{1}$ = 22.38 $\approx$ 22.4L = 22.4 $\times$ $10^{-3}$ = $2.24 \times 10^{-2} m^3$

2   2- Find the number of molecules of an ideal gas in a volume of 1.000 $cm^3$ at STP.

n = $\frac{PV}{RT}$ = $\frac{1\times1\times10^{-3}}{0.082\times273}$ = $\frac{1}{22400}$$\\No of molecules = 6.023 \times 10^{23} \times \frac{1}{22400} = 2.688 \times 10^{19} 3 3- Find the number of molecules in 1 cm^3 of an ideal gas at 0°C and at a pressure of 10^{-6} mm of mercury. ##### Solution : V = 1 cm^3, T = 0^{\circ}C, P = 10^{-5} mm of Hg//n = \frac{PV}{RT} = \frac{fgh\times V}{RT} = \frac{1.36\times 980\times 10^{-6}\times 1}{8.31\times 273} = 5.874 \times10^{-13}$$\\$ No. of molecules = $No\times n$ = $6.023\times 10^{23}\times 5.874\times10^{-13}$ = 3.538 $\times$ $10^{11}$

4   4- Calculate the mass of 1 $cm^3$ of oxygen kept at STP.

$n = \frac{PV}{RT} = \frac{1\times 1\times 10^{-3}}{0.082\times 273}$ = $\frac{10^{-3}}{22.4}$$\\mass = \frac{(10^{-3}\times 32)}{22.4}g = 1.428 \times 10^{-3} g = 1.428 mg 5 5- Equal masses of air are sealed in two vessels, one of volume V_0 and the other of volume 2V_0 . If the first vessel is maintained at a temperature 300 K and the other at 600 K, find the ratio of the pressures in the two vessels. ##### Solution : Since mass is same\\$$n_1$ = $n_2$ = n$\\$$P_1 = \frac{nR\times 300}{V_0}, P_2 = \frac{nR\times 600}{2V_0}$$\\$$\frac{P_1}{P_2} = \frac{nR\times 300}{V_0} \times \frac{2V_0}{nR\times 600} = \frac{1}{1} 6 6- An electric bulb of volume 250 cc was sealed during manufacturing at a pressure of 10^{-3} mm of mercury at 27°C. Compute the number of air molecules contained in the bulb. Avogadro constant = 6 \times 10^{23} per mol, density of mercury = 13600 \frac{kg}{m^3} and g = 10 \frac{m}{s^2} . ##### Solution : V = 250 cc = 250 \times 10^{-3}$$\\$P = $10^{-3}$mm = $10^{-3} \times 10^{-3}$ m = $10^{-6}$ $\times$ 13600 $\times$ 10 pascal = 136 $\times$ $10^{-3}$ pascal$\\$T = 27$^{\circ}$C = 300K$\\$n = $\frac{PV}{RT}$ = $\frac{136\times 10^{-3}\times 250}{8.3\times 300}$ $\times$ $10^{-3}$ = $\frac{136\times 250}{8.3\times 300}$$\times 10^{-6}$$\\$No. of molecules = $\frac{136\times 250}{8.3\times 300}$ $\times 10^{-6}\times 6\times 10^{23}$ = 81 $\times$ $10^{17}$ $\approx$ 0.8 $\times$ $10^{15}$

7   7- A gas cylinder has walls that can bear a maximum pressure of 1.0 $\times$ $10^6$ Pa. It contains a gas at 8.0 $\times$ $10^6$ Pa and 300 K. The cylinder is steadily heated. Neglecting any change in the volume, calculate the temperature at which the cylinder will break.

$P_1$ = 8.0 $\times$ $10^5$ Pa, $P_2$ = 1 $\times$ $10^6$ Pa, $T_1$ = 300 K, $T_2$ = ?$\\$Since, $V_1$ = $V_2$ = V$\\$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \Rightarrow \frac{8\times 10^5\times V}{300} = \frac{1\times 10^6\times 300}{8\times 10^5} = 375^{\circ} K 8 8- 2 g of hydrogen is sealed in a vessel of volume 0*02 m^3 and is maintained at 300 K. Calculate the pressure in the vessel. ##### Solution : m = 2 g, V = 0.02 m_3 = 0.02 \times 10^6 cc = 0.02 \times 10^3L m = 2 g, V = 0.02 m_3 = 0.02 \times 10^6 cc = 0.02 \times 10^3L, T = 300K, P = ?\\M = 2 g,\\PV = nRT \Rightarrow PV = \frac{m}{M} RT \Rightarrow P \times 20 = \frac{2}{2}$$\times 0.082\times 300$$\\$$\Rightarrow$ P = $\frac{0.082\times 300}{20}$ = 1.23 atm = 1.23$\times 10^5 pa$ $\approx$ 1.23$\times 10^5$ pa

9   9- The density of an ideal gas is 1.25 x $10^{-3}$ $\frac{g}{cm^3}$ at STP. Calculate the molecular weight of the gas .

P = $\frac{fRt}{P}$ = $\frac{m}{M} \times \frac{RT}{V}$ = $\frac{fRT}{M}$$\\f \rightarrow 1.25 \times 10^{-3} \frac{g}{cm^3}$$\\$R $\rightarrow$ 8.31 $\times$ $10^7$ $\frac{ert}{deg}{mole}$ T $\rightarrow$ 273 K$\\$$\Rightarrow M = \frac{fRT}{P} = \frac{1.25\times 10^{-3}\times 8.31\times 10^7\times 273}{13.6\times 980\times 76} = 0.002796 \times 10^4 \approx 28 \frac{g}{mol} 10 10- The temperature and pressure at Simla are 15.0°C and 72.0 cm of mercury and at Kalka these are 35.0°C and 76.0 cm of mercury. Find the ratio of air density at Kalka to the air density at Simla. ##### Solution : T at Simla = 15^{\circ}C = 15 + 273 = 288 K\\ P at Simla = 72cm = 72 \times 10^{-2}\times 13600 \times 9.8$$\\$T at Kalka = 35$^{\circ}$C = 35 + 273 = 308 K$\\$P at Kalka = 76 cm = 76 $\times 10^{-2} \times 13600 \times 9.8$$\\PV = nRT\\$$\Rightarrow$ PV = $\frac{m}{M}RT$ $\Rightarrow$ PM = $\frac{m}{V}$RT $\Rightarrow$ f = $\frac{PM}{RT}$$\\$$\frac{fSimla}{fKalka}$ = $\frac{P_{simla}\times M}{RT_{simla}}\times \frac{RT_{kalka}}{P_{kalka}\times M}$ = 1.013$\\$= $\frac{72\times 10^{-2}\times 13600\times 9.8\times 308}{288\times 76\times 10^{-2}\times 13600 \times 9.8}$ = $\frac{72\times 308}{76\times 288}$ = 1.013$\\$$\frac{fKalka}{fSimla} = \frac{1}{1.013} = 0.987 T at Simla = 15^{\circ}C = 15 + 273 = 288 K\\ P at Simla = 72cm = 72 \times 10^{-2}\times 13600 \times 9.8$$\\$T at Kalka = 35$^{\circ}$C = 35 + 273 = 308 K$\\$P at Kalka = 76 cm = 76 $\times 10^{-2} \times 13600 \times 9.8$$\\PV = nRT\\$$\Rightarrow$ PV = $\frac{m}{M}RT$ $\Rightarrow$ PM = $\frac{m}{V}$RT $\Rightarrow$ f = $\frac{PM}{RT}$$\\$$\frac{f{Simla}}{f{Kalka}}$ = $\frac{P_{simla} \times M}{RT_{simla}} \times \frac{RT_{kalka}}{P_{kalka} \times M}$$\\= \frac{72\times 10^{-2}\times 13600\times 9.8\times 308}{288\times 76\times 10^{-2}\times 13600\times 9.8} 11 1. Calculate the volume of 1 mole of an ideal gas at STP. ##### Solution : 12 2. Find the number of molecules of an ideal gas in a volume of 1 cm^{3} at STP. ##### Solution : 13 3.Find the number of molecules in 1 cm3 of an ideal gas at 0°C and at a pressure of 10 ~6 mm of mercury. ##### Solution : 14 4. Calculate the mass of 1 cm 3 of oxygen kept at STP. ##### Solution : 15 5. Equal masses of air are sealed in two vessels, one of volume V0 and the other of volume 2V0. If the first vessel is maintained at a temperature 300 K and the other at 600 K, find the ratio of the pressures in the two vessels. ##### Solution : 16 5. Equal masses of air are sealed in two vessels, one of volume V0 and the other of volume 2V0. If the first vessel is maintained at a temperature 300 K and the other at 600 K, find the ratio of the pressures in the two vessels. ##### Solution : 17 11- Figure (24-E1) shows a cylindrical tube with adiabatic walls and fitted with a diathermic separator. The separator can be slid in the tube by an external mechanism. An ideal gas is injected in the two sides at equal pressures and equal temperatures. The separator remains in equilibrium at the middle. It is now slid to a position where it divides the tube in the ratio of 1:3. Find the ratio of the pressures in the two parts'of the vessel. ##### Solution : n_1 = n_2 = n\\$$P_1$ = $\frac{nRT}{V}$, $P_2$ = $\frac{nRt}{3V}$$\\$$\frac{P_1}{P_2}$ = $\frac{nRT}{V}$ $\times$ $\frac{3V}{nRT}$ = 3 : 1

18   12- Find the rms speed of hydrogen molecules in a sample of hydrogen gas at 300 K. Find the temperature at which the rms speed is double the speed calculated in the previous part.

r.m.s velocity of hydrogen molecules = ?$\\$T = 300 K, R = 8.3, M = 2 g = 2 $\times$ $10^{-3}$ Kg$\\$C = $\sqrt{\frac{3RT}{M}}$ $\Rightarrow$ C = $\sqrt{\frac{3 \times 8.3 \times 300}{2 \times 10^{-3}}}$ = 1932.6 $\frac{m}{s}$ $\approx$ 1930 $\frac{m}{s}$$\\Let the temp. at which the C = 2 × 1932.6 is T\\2 \times 1932.6 = \sqrt{\frac{3\times 8.3\times T  }{2\times 10^{-3}}} = \Rightarrow (2 \times 1932 .6)^2 = \frac{3\times 8.3\times T}{2\times 10^{-3}}$$\\$$\Rightarrow T = 1199.98 \approx 1200 K. 19 13- A sample of 0.177 g of an ideal gas occupies 1000 cm^3 at STP. Calculate the rms speed of the gas molecules. ##### Solution : V_{rms} = \sqrt{\frac{3P}{f}} P = 10^5 Pa = 1 atm, f = \frac{1.77\times 10^{-4}}{10^{-3}}$$\\$ = $\sqrt{\frac{3\times 10^5\times 10^{-3}}{1.77\times 10^{-4}}}$ = 1301.8 $\approx$ 1302 $\frac{m}{s}$

20   14- The average translational kinetic energy of air molecules is 0.040 eV (1 eV = 1.6 x $10^{-19}$ J). Calculate the temperature of the air. Boltzmann constant k= 1.38 x $10^{-23}$ J/K.

##### Solution :

Agv. K.E. = $\frac{3}{2}$ KT$\\$$\frac{3}{2} KT = 0.04 \times 1.6\times 10^{-19}$$\Rightarrow$ $\frac{3}{2} \times 1.38\times 10^{-23}\times T = 0.04 \times 1.6 \times 10^{-19}$$\\$$\Rightarrow$ T = $\frac{2\times 0.04\times 1.6\times 10^{-19}}{3\times 1.38\times 10^{-23}}$ = 0.0309178 $\times$ $10^4$ = 309.178 $\approx$ 310 K

21   15- Consider a sample of oxygen at 300 K. Find the average time taken by a molecule to travel a distance equal to the diameter of the earth.

##### Solution :

$V_{avg}$ = $\sqrt{\frac{8RT}{\pi M}}$ = $\sqrt{\frac {8\times 8.3\times 300}{3.14\times 0.032}}$$\\T = \frac{Distance}{Speed} = \frac{6400000\times 2}{445.25} = 445.25\frac{m}{s}$$\\$ = $\frac{28747.83}{3600}$ km = 7.985 $\approx$ 8 hrs.

22   16- Find the average magnitude of linear momentum of a helium molecule in a sample of helium gas at 0°C. Mass of a helium molecule = 6.64 $\times 10^{-27}$ kg and Boltzmann constant = 138 $\times 10^{23}$J/K.

M = 4$\times 10^{-3}$ Kg$\\$$V_{avg} =\sqrt{\frac{8RT}{\pi M}} = \sqrt{\frac{8\times 8.3\times 273}{3.14\times 4\times 10^{-3}}} = 1201.35 \\ Momentum = M\times V_{avg} = 6.64 \times 10^{-27} \times 1201.35 = 7.97 \times 10^{-24} \approx 8 \times 10^{-24} Kg-\frac{m}{s} 23 17- The mean speed of the molecules of a hydrogen sample equals the mean speed of the molecules of a helium sample. Calculate the ratio of the temperature of the hydrogen sample to the temperature of the helium sample. ##### Solution : V_{avg} = \sqrt{\frac{8RT}{\pi M}} = \frac{8\times 8.3\times 300}{3.14\times 0.032}$$\\$Now, $\frac{8RT_1}{\pi \times 2}$ = $\frac{8RT_2}{\pi \times 4}$$\\ \frac{T_1}{T_2} = \frac{1}{2} 24 18- At what temperature the mean speed of the molecules of hydrogen gas equals the escape speed from the earth ? ##### Solution : Mean speed of the molecule = \sqrt{\frac{8RT}{\pi M}}$$\\$Escape velocity = $\sqrt{2gr}$$\\$$\sqrt{\frac{8RT}{\pi M}}$ = $\sqrt{2gr}$ $\Rightarrow$ $\frac{8RT}{\pi M}$ = 2gr$\Rightarrow$ T = $\frac{2\times 9.8\times 6400000\times 3.14\times 2\times 10^{-3}}{8\times 8.3}$ $\\$= 11863.9 $\approx$ 11800 $\frac{m}{s}$

25   19- Find the ratio of the mean speed of hydrogen molecules to the mean speed of nitrogen molecules in a sample containing a mixture of the two gases.

##### Solution :

$V_{avg}$ = $\sqrt{\frac{8RT}{\pi M}}$$\\$$\frac{V_{avg}H_2}{V_{avg}N_2}$ = $\sqrt{\frac{8RT}{\pi \times 2}}$ $\times \sqrt{\frac{\pi \times 28}{8RT}}$ = $\sqrt{\frac{28}{2}}$ = $\sqrt{14}$ = 3.74

26   20- Figure (24-E2) shows a vessel partitioned by a fixed diathermic separator. Different ideal gases are filled in the two parts. The rms speed of the molecules in the left part equals the mean speed of the molecules in the right part. Calculate the ratio of the mass of a molecule in the left part to the mass of a molecule in the right part.

The left side of the container has a gas, let having molecular wt. $M_1$$\\Right part has Mol. wt = M_2$$\\$Temperature of both left and right chambers are equal as the separating wall is diathermic$\\$$\sqrt{\frac{3RT}{M_1}} = \sqrt{\frac{8RT}{\pi M_2}} \Rightarrow \frac{3RT}{M_1} = \frac{8RT}{M_1} \Rightarrow \frac{M_1}{\pi M_2} = \frac{3}{8} \Rightarrow \frac{M_1}{M_2} = \frac{3\pi }{8} \\= 1.1775 \approx 1.18 27 21- Estimate the number of collisions per second suffered by a molecule in a sample of hydrogen at STP. The mean free path (average distance covered by a molecule between successive collisions) = 1.38\times 10^{-7} cm. ##### Solution : V_{mean} = \sqrt{\frac{8RT}{\pi M}} = \sqrt{\frac{8\times 8.3\times 273}{3.14\times 2\times 10^{-3}}} = 1698.96\\ Total Dist = 1698.96 m\\No. of Collisions = \frac{1698.96}{1.38\times 10^{-7}} = 1.23\times 10^{10} 28 22- Hydrogen gas is contained in a closed vessel at 1 atm (100 kPa) and 300 K. (a) Calculate the mean speed of the molecules, (b) Suppose the molecules strike the wall with this speed making an average angle of 45° with it. How many molecules strike each square metre of the wall per second ? ##### Solution : P = 1 atm = 10^5 Pascal\\T = 300K, M = 2 g = 2 \times 10^{-3} Kg\\(a) V_{avg} = \sqrt{\frac{8RT}{\pi M}} = \sqrt{\frac{8\times 8.3\times 300}{3.14\times 2\times 10^{-3}}} = 1781.004 \approx 1780 \frac{m}{s}$$\\$(b) When the molecules strike at an angle 45°,$\\$Force exerted = mV Cos 45° - (-mV Cos 45°) = 2 mV Cos 45° = 2 m V $\frac{1}{2}$ = $\sqrt{2}mV$$\\No. of molecules striking per unit area = \frac{Force}{\sqrt{2}mv \times Area} = \frac{Pressure}{\sqrt{2}mV}$$\\$= $\frac{10^5\times 6\times 10^{23}}{\sqrt2\times 2\times 10^{-3}\times 1780}$ = $\frac{3}{\sqrt2\times 1780}\times 10^{31}$ = 1.19 $\times 10^3\times 10^{31}$ $\\$= 1.19 $\times 10^{28} \approx 1.2\times 10^{28}$

29   23- Air is pumped into an automobile tyre's tube upto a pressure of 200 kPa in the morning when the air temperature is 20°C. During the day the temperature rises to 40°C and the tube expands by 2%. Calculate the pressure of the air in the tube at this temperature.

##### Solution :

$\frac{P_1 V_1}{T_1}$ = $\frac{P_2 V_2}{T_2}$$\\$$P_1$ $\rightarrow$ 200 KPa = 2 $\times 10^5$ pa $P_2$ = ?$\\$$T_1 = 20^\circ = 293 K T_2 = 40^\circC = 313K \\$$V_2$ = $V_1 + 2\% V_1$ = $\frac{102\times V_1}{100}$$\\$$\Rightarrow$ $\frac{2\times 10^5\times V_1}{293}$ = $\frac{P_2\times 102\times V_1}{100\times 313}$ $\Rightarrow$ $P_2$ = $\frac{2\times 10^7\times 313}{102\times 293}$ = 209462 Pa $\\$= 209.462Kpa

30   24- Oxygen is filled in a closed metal jar of volume 1.0 $\times 10^{-3} m^3$ at a pressure of 1.5 $\times 10^5$ Pa and temperature 400 K. The jar has a small leak in it. The atomospheric pressure is1.0 $\times 10^5$ Pa and the atmospheric temperature is 300 K. Find the mass of the gas that leaks out by the time the pressure and the temperature inside the jar equalise with the surrounding.

$V_1$ = 1 $\times 10^{-3} m^3$, $P_1$ = 1.5$\times 10^5$ Pa, $T_1$ = 400 K$\\$$P_1 V_1 = n_1R_1T_1$$\\$$\Rightarrow n = \frac{P_1V_1}{R_1T_1} = \frac{1.5\times 10^5\times 1\times 10^{-3}}{8.3\times 400} \Rightarrow n = \frac{1.5}{8.3\times 4}$$\\$$\Rightarrow m_1 = \frac{1.5}{8.3\times 4}\times 32 = 1.4457 \approx 1.446$$\\$$P_2 = 1\times 10^5 Pa, V_2 = 1\times 10^{-3} m^{3}, T_2 = 300K, T_2 = 300 K\\$$P_2V_2 = n_2R_2T_2$$\\$$\Rightarrow n_2 = \frac{P_2V_2}{R_2T_2}$ = $\frac{10^5\times 10^{-3}}{8.3\times 300}$ = $\frac{1}{3\times 8.3}$ = 0.040$\\$$\Rightarrow m_2 = 0.4 \times 32 = 1.285$$\\$$\Delta m = m_1 - m_2 = 1.446 - 1.285 = 0.1608 g \approx 0.16 g 31 25- An air bubble of radius 2.0 mm is formed at the bottom of a 3.3 m deep river. Calculate the radius of the bubble as it comes to the surface. Atmospheric pressure = 1.0 x 10^5 Pa and density of water = 1000 kg/m^3 . ##### Solution : P_1 = 10^5 + ƒgh = 10^5 + 1000 × 10 × 3.3 = 1.33 × 10^5 pa\\$$P_2$ = $10^5$, $T_1$ = $T_2$ = T, $V_1$ = $\frac{4}{3} \pi(2 \times 10^{-3})^{-3}$$\\$$V_2$ = $\frac{4}{3} \pi r^3$, r = ?$\\$ $\frac{P_1 V_1}{T_1}$ = $\frac{P_2 V_2}{T_2}$$\\ \Rightarrow n = \frac{1.33\times 8\times 10^5\times \frac{4}{3} \times \pi \times (2\times 10^{-3})^3}{T_1} = \frac{10^5\times \frac{4}{3}\times \pi r^2}{T_2}$$\Rightarrow 1.33 \times 8\times 10^5\times 10^{-9} = 10^5\times r^3$ $\\$ $\Rightarrow$ r = $\sqrt[3]{10.64\times 10^{-3}}$ = 2.19 $\times$ $10^{-3}$ $\approx$ 2.2 mm

32   26- Air is pumped into the tubes of a cycle rickshaw at a pressure of 2 atm. The volume of each tube at this pressure is 0.002 $m^3$ . One of the tubes gets punctured and the volume of the tube reduces to 0.0005 $m^3$ . How many moles of air have leaked out ? Assume that the temperature remains constant at 300 K and that the air behaves as an ideal gas.

$P_1$ = 2 atm = 2$\times 10^5$ pa$\\$$V_1 = 0.002 m^3, T_1 = 300K\\$$P_1 V_1$ = $n_1RT_1$$\\$$\Rightarrow$ n = $\frac{P_1V_1}{RT_1}$ = $\frac{2\times 10^5\times 0.002}{8.3\times 300}$ = $\frac{4}{8.3\times 3}$ = 0.1606$\\$$P_2 = 1 atm = 10^5 pa \\$$V_2$ = 0.0005 $m^3$, $T_2$ = 300 K$\\$$P_2 V_2 = n_2RT_2$$\\$$\Rightarrow n_2 = \frac{P_2V_2}{RT_2} = \frac{5}{3\times 8.3}$$\times$ $\frac{1}{10}$ = 0.02$\\$$\Deltan = moles leaked out = 0.16 - 0.02 = 0.14 33 27- 0.040 g of He is kept in a closed container initially at 100.0°C. The container is now heated. Neglecting the expansion of the container, calculate the temperature at which the internal energy is increased by 12 J. ##### Solution : m = 0.040g, T = 100°C, M_{He} = 4 g\\U = \frac{3}{2}nRT = \frac{3}{2} \frac{m}{M} \times RT'$$\\$$\Rightarrow 1.5 \times0.01\times 8.3\times 373 + 12 = 1.5\times 0.01\times 8.3\times T'\Rightarrow T' = \frac{58.4385}{0.1245} = 469.3855 K = 196.3°C \approx 196°C 34 28- During an experiment, an ideal gas is found to obey an additional law pV^2 \rightarrow constant. The gas is initially at a temperature T and volume V. Find the temperature when it expands to a volume 2V. ##### Solution : PV^2 = constant\\$$\Rightarrow$ $P_1V_1^2$ = $P_2V_2^2$$\\$$\Rightarrow$ $\frac{nRT_1}{V_1}\times {V_1^2}$ = $\frac{nRT_2}{V_2}\times V_2^2$$\\$$\Rightarrow$ $T_1 V_1 = T_2 V_2 = TV = T_1 \times 2V$ $\Rightarrow T_2 = \frac{T}{2}$

35   29- A vessel contains 1.60 g of oxygen and 2.80 g of nitrogen. The temperature is maintained at 300 K and the volume of the vessel is 0.166$m^3$ . Find the pressure of the mixture.

$P_{O_2} = \frac{n_{O_2} RT}{V}$, $P_{H_2} = \frac{n_{H_2}RT}{V}$$\\$$n_{O_2} = \frac{m}{M_{O_2}} = \frac{1.60}{32} = 0.05$$\\Now, P_{mix} = (\frac {n_{O_2} + n_{H_2}}{V})RT$$\\$$n_{H_2} = \frac{m}{M_{H_2}} = \frac{2.80}{28} = 0.1$$\\$$P_{mix} = \frac{(0.05 + 0.1)\times 8.3\times 300}{0.166} = 2250 \frac{N}{m^2} 36 30- A vertical cylinder of height 100 cm contains air at a constant temperature. The top is closed by a frictionless light piston. The atmospheric pressure is equal to 75 cm of mercury. Mercury is slowly poured over the piston. Find the maximum height of the mercury column that can be put on the piston. ##### Solution : P_1 = Atmospheric pressure = 75 × ƒg\\$$V_1$ = 100 × A$\\$$P_2 = Atmospheric pressure + Mercury pessue = 75ƒg + hgƒg (if h = height of mercury)\\$$V_2$ = (100 - h) A$\\$$P_1V_1 = P_2V_2$$\\$$\Rightarrow 75ƒg(100A) = (75 + h)ƒg(100 – h)A\\$$\Rightarrow$ 75 × 100 = (74 + h) (100 – h) $\Rightarrow$ 7500 = 7500 – 75 h + 100 h –$h^2$$\\$$\Rightarrow$ $h^2$ - 25 h = 0 $\Rightarrow$ $h^2$ = 25 h$\Rightarrow$ h = 25cm$\\$$\RightarrowHeight of mercury that can be poured = 25 cm 37 31- Figure (24-E3) shows two vessels A and B with rigid walls containing ideal gases. The pressure, temperature and the volume are p_A , T_A , V in the vessel A and P_B, T_B , V in the vessel B. The vessels are now connected through a small tube. Show that the pressure p and the temperature T satisfy\\$$\frac{p}{T} = \frac{1}{2}(\frac{p_A}{T_A} + \frac{p_B}{T_B})$$\\when equilibrium is achieved. ##### Solution : Now, Let the final pressure; Volume & Temp be\\After connection = P_A' \rightarrow Partial pressure of A\\$$P_B' \rightarrow$ Partial pressure of B$\\$Now, $\frac{P_A' \times 2V}{T} = \frac{P_A \times V}{T_A}$$\\Or \frac{P_A'}{T} = \frac{P_A \times V}{T_A} ...(1)\\Similarly, \frac{P_B'}{T} = \frac{P_B}{2T_B} ...(2)\\Adding (1) & (2)\\$$\frac{P_A'}{T} + \frac{P_B'}{T} = \frac{1}{2}(\frac{P_A}{T_A} + \frac{P_B}{T_B})$$\\$$\Rightarrow$ $\frac{P}{T} = \frac{1}{2}(\frac{P_A}{T_A} + \frac{P_B}{T_B})$ [$\therefore$ $P_A' + P_B' = P$]

38   32- A container of volume 50 cc contains air (mean molecular weight = 28.8 g) and is open to atmosphere where the pressure is 100 kPa. The container is kept in a bath containing melting ice (0°C). (a) Find the mass of the air in the container when thermal equilibrium is reached, (b) The container is now placed in another bath containing boiling water (100°C). Find the mass of air in the container, (c) The container is now closed and placed in the melting-ice bath. Find the pressure of the air when thermal equilibrium is reached.

V = 50 cc = 50 $\times 10^{-6} cm^3$$\\P = 100 KPa = 10^5 Pa M = 28.8 g\\(a) PV = nr T_1$$\\$$\Rightarrow PV = \frac{m}{M}RT_1 \Rightarrow m = \frac{PMV}{RT_1} = \frac{10^5 \times 28.8\times 50\times 10^{-6}}{8.3\times 273} =\\ \frac{50\times 28.8\times 10^{-1}}{8.3\times 273} = 0.0635 g\\(b) When the vessel is kept on boiling water\\PV = \frac{m}{M}RT_2$$\Rightarrow m = \frac{PVM}{RT_2}$ = $\frac{10^5\times 28.8\times 50\times 10^{-6}}{8.3\times 373}$ = $\frac{50\times 28.8\times 10^{-1}}{8.3\times 373}$ = 0.0465$\\$(c) When the vessel is closed$\\$$P\times 50\times 10^{-6} = \frac{0.0465}{28.8}\times 8.3\times 273$$\\$$\Rightarrow P = \frac{0.0465\times 8.3\times 273}{28.8\times 50\times 10^{-6}} = 0.07316\times 10^6 Pa \approx 73 KPa 39 33- A uniform tube closed at one end, contains a pallet of mercury 10 cm long. When the tube is kept vertically with the closed end upward, the length of the air column trapped is 20 cm. Find the length of the air column trapped when the tube is inverted so that the closed end goes down. Atmospheric pressure = 75 cm of mercury. ##### Solution : Case I \rightarrow Net pressure on air in volume V\\$$ = P_{atm} - hƒg = 75 × ƒ_{Hg} - 10 ƒ_{Hg} = 65 × ƒ_{Hg} × g$$\\Case II \rightarrow Net pressure on air in volume V = P_{atm} + f_{Hg} × g × h$$\\$$P_1 V_1 = P_2 V_2 \\$$\Rightarrow f_{Hg}\times g\times 65\times A\times 20 = f_{Hg}\times g\times 75 + f_{Hg}\times g\times 10\times A\times h$$\\$$\Rightarrow 62\times 20 = 85 h \Rightarrow h = \frac{65\times 20}{85}$ = 15.2 cm$\approx$ 15 cm

40   34- A glass tube, sealed at both ends, is 100 cm long. It lies horizontally with the middle 10 cm containing mercury. The two ends of the tube contain air at 27°C and at a pressure 76 cm of mercury. The air column on one side is maintained at 0°C and the other side is maintained at 127°C. Calculate the length of the air column on the cooler side. Neglect the changes in the volume of mercury and of the glass.

##### Solution :

2L + 10 = 100 $\Rightarrow$ 2L = 90 $\Rightarrow$ L = 45 cm$\\$Applying combined gas eqn to part 1 of the tube$\\$$\frac{(45A)P_0}{300} = \frac{(45 - x)P_1}{273}$$\\$$\Rightarrow P_1 = \frac{273\times 45\times P_0}{300(45 - x)}$$\\$

Applying combined gas eqn to part 2 of the tube$\\$$\frac{45AP_0}{300} = \frac{(45 + x)AP_2}{400}$$\\$$\Rightarrow P_2 = \frac{400\times 45\times P_0}{300(45\times - x)}$$\\$$P_1 = P_2$$\\$$\Rightarrow \frac{273\times 45\times P_0}{300(45 - x)} = \frac{400\times 45\times P_0}{300(45 + x)}$$\\$$\Rightarrow (45 - x) 400 = (45 + x) 273 \Rightarrow 18000 - 400 x = 12285 + 273 x$$\\$$\Rightarrow (400 + 273)x = 18000 – 12285 \Rightarrow x = 8.49$$\\$$P_1 = \frac{273\times 46\times 76}{300\times 36.51} = 85 % 25 cm of Hg$$\\$Length of air column on the cooler side = L - x = 45 - 8.49 = 36.51

41   35- An ideal gas is trapped between a mercury column and the closed end of a narrow vertical tube of uniform base containing the column. The upper end of the tube is open to the atmosphere. The atmospheric pressure equals 76 cm of mercury. The lengths of the mercury column and the trapped air column are 20 cm and 43 cm respectively. What will be the length of the air column when the tube is tilted slowly in a vertical plane through an angle of 60° ? Assume the temperature to remain constant.

44   38- One mole of an ideal gas undergoes a process$\\$p = $\frac{p_0}{1 + (\frac{V}{V_0})^2}$$\\where p_0 and V_0 are constants. Find the temperature of the gas when V = V_0 . ##### Solution : P = \frac{P_0}{1 + (\frac{V}{V_0})^2}$$\\$$\Rightarrow \frac{nRT}{V} = \frac{P_0}{1 + (\frac{V}{V_0})^2} [PV = nRT according to ideal gas equation]\\$$\Rightarrow \frac{RT}{V} = \frac{P_0}{1+(\frac{V}{V_0})_2}$ [Since n = 1 mole]$\\$$\Rightarrow \frac{RT}{V_0} = \frac{P_0}{1+(\frac{V}{V_0})_2} [At V = V_0 ]\\$$\Rightarrow P_0V_0 = RT(1+1) \Rightarrow P_0V_0 = 2 RT \Rightarrow T = \frac{P_0V_0}{2R}$

45   39- Show that the internal energy of the air (treated as an ideal gas) contained in a room remains constant as the temperature changes between day and night. Assume that the atmospheric pressure around remains constant and the air in the room maintains this pressure by communicating with the surrounding through the windows etc.

Internal energy = nRT$\\$Now, PV = nRT$\\$$nT = \frac{PV}{R} Here P & V constant\\$$\Rightarrow$ nT is constant$\\$$\therefore Internal energy = R × Constant = Constant 46 40- Figure (24-E5) shows a cylindrical tube of radius 5 cm and length 20 cm. It is closed by a tight-fitting cork. The friction coefficient between the cork and the tube is 0.20. The tube contains an ideal gas at a pressure of 1 atm and a temperature of 300 K. The tube is slowly heated and it is found that the cork pops out when the temperature reaches 600 K. Let dN denote the magnitude of the normal contact force exerted by a small length dl of the cork along the periphery (see the figure). Assuming that the temperature of the gas is uniform at any instant, calculate \frac{dN}{dT}. ##### Solution : Frictional force = \mu N\\Let the cork moves to a distance = dl\\$$\therefore$ Work done by frictional force = $\mu Nde$$\\Before that the work will not start that means volume remains constant\\$$\Rightarrow \frac{P_1}{T_1} = \frac{P_2}{T_2} \Rightarrow \frac{1}{300} = \frac{P_2}{600} \Rightarrow P_2 = 2 atm$$\\$$\therefore$ Extra Pressure = 2 atm - 1 atm = 1 atm$\\$Work done by cork = 1 atm (Adl) $\mu$ Ndl = [1atm][Adl]$\\$$N = \frac{1\times 10^5\times (5\times 10^{-2})^2}{2} = \frac{1\times 10^5\times \pi\times 25\times 10^{-5}}{2}$$\\$Total circumference of work = $2\pi r \frac{dN}{dl} = \frac{N}{2\pi r}$$\\= \frac{1\times 10^5\times \pi \times 25\times 10^{-5}}{0.2\times 2\pi r} = \frac{1\times 10^5\times 25\times 10^{-5}}{0.2\times 2\times 5\times 10^5} = 1.25\times 10^4 \frac{N}{M} 47 41- Figure (24-E6) shows a cylindrical tube of cross-sectional area A fitted with two frictionless pistons. The pistons are connected to each other by a metallic wire. Initially, pressure inside the bulbs ? The volume of the connecting tube is negligible. ##### Solution : \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$\\$$\Rightarrow \frac{P_0 V}{T_0} = \frac{P' V}{2T_0} \Rightarrow P' = 2 P_0$$\\$Net pressure = $P_0$ outwards $\\$$\therefore Tension in wire = P_0 A$$\\$Where A is area of tube.

48   42- Figure (24-E7) shows a large closed cylindrical tank containing water. Initially the air trapped above the water surface has a height $h_0$ and pressure $2p_0$ where $p_0$ is the atmospheric pressure. There is a hole in the wall of the tank at a depth $h_1$ below the top from which water comes out. A long vertical tube is connected as shown, (a) Find the height $h_2$ of the water in the long tube above the top initially, (b) Find the speed with which water comes out of the hole.(c) Find the height of the water in the long tube above the top when the water stops coming out of the hole.

##### Solution :

(a) $2P_0x = (h_2 + h_0)fg$$\\$$\Rightarrow 2P_0 = h_2 fg + h_0 fg$$\\$$\Rightarrow h_2 fg = 2P_0 - h_0 fg$$\\$$h_2 = \frac{2P_0}{fg} - \frac{h_0 fg}{fg} = \frac{2P_0}{fg} - h_0$$\\(b) K.E. of the water = Pressure energy of the water at that layer\\$$\Rightarrow \frac{1}{2}mV^2 = m\times \frac{P}{f}$$\\ (a) 2P_0x = (h_2 + h_0)fg$$\\$$\Rightarrow 2P_0 = h_2 fg + h_0 fg$$\\$$\Rightarrow h_2 fg = 2P_0 - h_0 fg$$\\$$h_2 = \frac{2P_0}{fg} - \frac{h_0 fg}{fg} = \frac{2P_0}{fg} - h_0$$\\$(b) K.E. of the water = Pressure energy of the water at that layer$\\$$\Rightarrow \frac{1}{2}mV^2 = m\times \frac{P}{f}$$\\$$\Rightarrow V^2 = \frac{2P}{f} = [\frac{2}{f(P_0 + fg(h_1 - h_0)}]$$\\$$\Rightarrow V = [\frac{2}{f(P_0 + fg(h_1 - h_0)}]^\frac{1}{2}$$\\$(c) $(x + P_0 )ƒh = 2P_0$$\\$$\therefore 2P_0 + ƒg (h -h_0)= P_0 + ƒgx$$\\$$\therefore X = \frac{P_0}{fg+h_1-h_0} = h_2 + h_1$$\\$$\therefore$ i.e. x is $h_1$ meter below the top $\Rightarrow$ x is $-h_1$ above the top

49   43- An ideal gas is kept in a long cylindrical vessel fitted with a frictionless piston of cross-sectional area $10 cm^2$ and weight 1 kg (figure 24-E8). The vessel itself is kept in a big chamber containing air at atmospheric pressure 100 kPa. The length of the gas column is 20 cm. If the chamber is now completely evacuated by an exhaust pump, what will be the length of the gas column ? Assume the temperature to remain constant throughout the process.

##### Solution :

50   44- An ideal gas is kept in a long cylindrical vessel fitted with a frictionless piston of cross-sectional area $10cm^2$ and weight 1 kg. The length of the gas column in the vessel is 20 cm. The atmospheric pressure is 100 kPa. The vessel is now taken into a spaceship revolving round the earth as a satellite. The air pressure in the spaceship is maintained at 100 kPa. Find the length of the gas column in the cylinder.

##### Solution :

$P_1 V_1 = P_2 V_2$$\\$$\Rightarrow (\frac{mg}{A} + P_0)Al P_0 Al$$\\$$\Rightarrow (\frac{1\times 9.8}{10\times 10^{-4}} + 10^5)0.2 = 10^5 l'$$\\$$\Rightarrow (9.8\times 10^3 + 10^5)\times 0.2 = 10^5 l'$$\\$$\Rightarrow 109.8\times 10^3\times 0.2 = 10^5 l'$$\\$$\Rightarrow l' = \frac{109.8\times 0.2}{10^2} = 0.2196 \approx 0.22m\approx 22cm$

51   45- Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at 0°C at a pressure of 76 cm of mercury. One of the bulbs is then placed in melting ice and the other is placed in a water bath maintained at 62°C. What is the new value of the pressure inside the bulbs ? The volume of the connecting tube is negligible.

When the bulbs are maintained at two different temperatures.$\\$The total heat gained by ‘B’ is the heat lost by ‘A’$\\$Let the final temp be x so, $m_1 S\Delta t = m_2 S\Delta t$$\Rightarrow n_1 M\times s(x-0) = n_2 M\times S\times (62 - x) \Rightarrow n_1 x = 62n_2 - n_2 x$$\\$$\Rightarrow x = \frac{62n_2}{n_1 + n_2} = \frac{62n_2}{2n_2} = 31°C = 304 K$$\\$For a single ball, Initial Temp = 0°C, P = 76 cm of Hg$\\$$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}, V_1 = V_2, Hence n_1 = n_2$$\\$$\Rightarrow \frac{76\times V}{273} = \frac{P_2\times V}{304} \Rightarrow P_2 = \frac{403\times 76}{273} = 84.630 \approx84°C 52 46- The weather report reads, "Temperature 20°C : Relative humidity 100%. What is the dew point ? ##### Solution : Temp is 20° , Relative humidity = 100%\\So the air is saturated at 20°C\\Dew point is the temperature at which SVP is equal to present vapour pressure\\So 20°C is the dew point. 53 47- The condition of air in a closed room is described as follows. Temperature - 25°C, relative humidity = 60%, pressure = 104 kPa. If all the water vapour is removed from the room without changing the temperature, what will be the new pressure ? The saturation vapour pressure at 25°C = 3.2 kPa. ##### Solution : T = 25°C P = 104 KPa\\$$RH = \frac{VP}{SVP} , [SVP = 3.2 KPa , RH = 0.6]$$\\VP = 0.6 \times 3.2 \times 10 = 1.92 \times 10 \approx 2 \times 10^3$$\\$When vapours are removed VP reduces to zero$\\$Net pressure inside the room now = $104 \times 10^3 - 2 \times 10^3 = 102 \times 10^3 = 102 KPa$

54   48- The temperature and the dew point in an open room are 20°C and 10°C. If the room temperature drops to 15°C, what will be the new dew point ?

##### Solution :

Temp = 20°C , Dew point = 10°C$\\$The place is saturated at 10°C$\\$Even if the temp drop dew point remains unaffected.$\\$The air has V.P. which is the saturation VP at 10°C. It (SVP) does not change on temp.

55   49- Pure water vapour is trapped in a vessel of volume 10 cm . The relative humidity is 40%. The vapour is compressed slowly and isothermally. Find the volume of the vapour at which it will start condensing.

$RH = \frac{VP}{SVP}$$\\The point where the vapour starts condensing, VP = SVP\\We know P_1 V_1 = P_2 V_2$$\\$$R_H SVP \times 10 = SVP \times V_2 \Rightarrow V_2 = 10R_H \Rightarrow 10\times 0.4 = 4 cm^3 56 50- A barometer tube is 80 cm long (above the mercury reservoir). It reads 76 cm on a particular day. A small amount of water is introduced in the tube and the reading drops to 75.4 cm. Find the relative humidity in the space above the mercury column if the saturation vapour pressure at the room temperature is l.0 cm. ##### Solution : Atm-Pressure = 76 cm of Hg\\When water is introduced the water vapour exerts some pressure which counter acts the atm pressure.\\The pressure drops to 75.4 cm\\Pressure of Vapour = (76 - 75.4) cm = 0.6 cm\\R. Humidity = \frac{VP}{SVP} = \frac{0.6}{1} = 0.6 = 60\% 57 51- Using figure (24.6) of the text, find the boiling point of methyl alcohol at 1 atm (760 mm of mercury) and at 0.5 atm. ##### Solution : From fig. 24.6, we draw \perpr, from Y axis to meet the graphs.\\Hence we find the temp. to be approximately 65°C & 45°C 58 52- The human body has an average temperature of 98 °F. Assume that the vapour pressure of the blood in the veins behaves like that of pure water. Find the minimum atmospheric pressure which is necessary to prevent the blood from boiling. Use figure (24.6) of the text for the vapour pressures. ##### Solution : The temp. of body is 98°F = 37°C\\At 37°C from the graph SVP = Just less than 50 mm\\ B.P. is the temp. when atmospheric pressure equals the atmospheric pressure.\\ Thus min. pressure to prevent boiling is 50 mm of Hg. 59 53- A glass contains some water at room temperature 20°C. Refrigerated water is added to it slowly. When the temperature of the glass reaches 10°C, small droplets condense on the outer surface. Calculate the relative humidity in the room. The boiling point of water at a pressure of 17.5 mm of mercury is 20°C and at 8.9 mm of mercury it is 10°C. ##### Solution : Given\\ SVP at the dew point = 8.9 mm , SVP at room temp = 17.5 mm\\ Dew point = 10°C as at this temp. the condensation starts \\Room temp = 20°C\\$$RH = \frac{SVP at dew point}{SVP at room temp} = \frac{8.9}{17.5} = 0.508 \approx 51\%$

60   54- 50 $cm^3$ of saturated vapour is cooled down from 30°C to 20°C. Find the mass of the water condensed. The absolute humidity of saturated water vapour is 30$\frac{g}{m^3}$ at 30°C and 16$\frac{g}{m^3}$ at 20°C.

##### Solution :

50 $cm^3$ of saturated vapour is cooled 30° to 20°. The absolute humidity of saturated $H_{2}O$ vapour 30 $\frac{g}{m^3}$ $\\$Absolute humidity is the mass of water vapour present in a given volume at 30°C, it contains 30 $\frac{g}{m^3}$ at 50 $m^3$ it contains 30 × 50 = 1500 g$\\$ at 20°C it contains 16 × 50 = 800 g$\\$ Water condense = 1500 – 800 = 700 g.

61   55- A barometer correctly reads the atmospheric pressure as 76 cm of mercury. Water droplets are slowly introduced into the barometer tube by a dropper. The height of the mercury column first decreases and then becomes constant. If the saturation vapour pressure at the atmospheric temperature is 0'80 cm of mercury, find the height of the mercury column when it reaches its minimum value.

##### Solution :

Pressure is minimum when the vapour present inside are at saturation vapour pressure As this is the max. pressure which the vapours can exert.$\\$ Hence the normal level of mercury drops down by 0.80 cm $\therefore$The height of the Hg column = 76 – 0.80 cm = 75.2 cm of Hg.$\\$ [$\therefore$ Given SVP at atmospheric temp = 0.80 cm of Hg]

62   56- 50 cc of oxygen is collected in an inverted gas jar over water. The atmospheric pressure is 99.4 kPa and the room temperature is 27°C. The water level in the jar is same as the level outside. The saturation vapour pressure at 27°C is 3.4 kPa. Calculate the number of moles of oxygen collected in the jar.

Pressure inside the tube = Atmospheric Pressure = 99.4 KPa$\\$ Pressure exerted by $O_2$ vapour= Atmospheric pressure - V.P.$\\$ = 99.4 KPa – 3.4 KPa = 96 KPa$\\$No of moles of $O_2$ = n$\\$96 × $10^3$ ×50 × $10^{-6}$ = n × 8.3 × 300$\\$$\Rightarrow n = \frac{96\times 50\times 10^{-3}}{8.3\times 300} = 1.9277 \times 10^{-3} \approx 1.93\times 10^{-3} 63 57- A faulty barometer contains certain amount of air and saturated water vapour. It reads 74.0 cm when the atmospheric pressure is 76.0 cm of mercury and reads 72.10 cm when the atmospheric pressure is 74.0 cm of mercury. Saturation vapour pressure at the air temperature = 1.0 cm of mercury. Find the length of the barometer tube above the mercury level in the reservoir. ##### Solution : Let the barometer has a length = x\\Height of air above the mercury column = (x – 74 – 1) = (x – 73)\\ Pressure of air = 76 – 74 – 1 = 1 cm\\For 2^{nd} case height of air above = (x – 72.1 – 1 – 1) = (x – 71.1)\\Pressure of air = (74 – 72.1 – 1) = 0.99\\$$(x - 73)(1) = \frac{9}{10}(x - 71.1)$ $\Rightarrow 10(x-73) = 9 (x-71.1)$$\\$$\Rightarrow x = 10\times 73 - 9\times 71.1 = 730 - 639.9 = 90.1$$\\Height of air = 90.1\\Height of barometer tube above the mercury column = 90.1 + 1 = 91.1 mm 64 58- On a winter day, the outside temperature is 0°C and relative humidity 40%. The air from outside comes into a room and is heated to 20°C. What is the relative humidity in the room ? The saturation vapour pressure at 0°C is 4.6 mm of mercury and at 20°C it is 18 mm of mercury. ##### Solution : Relative humidity = 40%\\SVP = 4.6 mm of Hg\\0.4 = \frac{VP}{4.6} \Rightarrow VP = 0.4\times 4.6 = 1.84$$\\$$\frac{P_1 V}{T_1} = \frac{P_2 V}{T_2} \Rightarrow \frac{1.84}{273} = \frac{P_2}{293} \Rightarrow P_2 = \frac{1.84}{273}\times 293Relative humidity at 20°C\\= \frac{VP}{SVP} = \frac{1.84\times 293}{273\times 10} = 0.109 = 10.9\% 65 59- The temperature and humidity of air are 27°C and 50% on a particular day. Calculate the amount of vapour that should be added to 1 cubic metre of air to saturate it. The saturation vapour pressure at 27°C = 3600 Pa. ##### Solution : RH = \frac{VP}{SVP}$$\\$Given, $0.50 = \frac{VP}{3600}$$\Rightarrow VP = 3600\times 0.5$$\\$Let the Extra pressure needed be P$\\$So, $P = \frac{m}{M}\times \frac{RT}{V} = \frac{m}{18}\times \frac{8.3\times 300}{1}$$\\Now, \frac{m}{18}\times 8.3\times 300 + 3600\times 0.50 = 3600 [air is saturated i.e. RH = 100% = 1 or VP = SVP]\\$$\Rightarrow m =(\frac{36 - 18}{8.3})\times 6 = 13 g$

66   60- The temperature and relative humidity in a room are 300 K and 20% respectively. The volume of the room is 50 $m^3$ . The saturation vapour pressure at 300 K is 3.3 kPa. Calculate the mass of the water vapour present in the room.

T = 300 K, Rel. humidity = 20%, V = 50 $m^3$$\\SVP at 300 K = 3.3 KPa, V.P. = Relative humidity × SVP = 0.2 × 3.3 × 10^3$$\\$$PV = \frac{m}{M}RT\Rightarrow 0.2× 3.3 ×$$10^3$ × 50 = $\frac{m}{18}\times 8.3\times 300$$\\$$\Rightarrow m = \frac{0.2\times 3.3\times 50\times 18\times 10^3}{8.3\times 300}$ = 238.55 grams $\approx 238g$$\\Mass of water present in the room = 238 g. 67 61- The temperature and the relative humidity are 300 K and 20% in a room of volume 50m^3 . The floor is washed with water, 500 g of water sticking on the floor. Assuming no communication with the surrounding, find the relative humidity when the floor dries. The changes in temperature and pressure may be neglected. Saturation vapour pressure at 300 K = 3.3 kPa. ##### Solution : RH = \frac{VP}{SVP} \Rightarrow 0.20 = \frac{VP}{3.3\times 10^3} = 660$$\\$$PV = nRT\Rightarrow P = \frac{nRT}{V} = \frac{m}{V}\times \frac{RT}{V} = \frac{500}{18}\times \frac{8.3\times 300}{50} = 1383.3$$\\$Net P = 1383.3 + 660 = 2043.3$\\$Now RH, = $\frac{2034.3}{3300} = 0.619 \approx 62\%$

68   62- A bucket full of water is placed in a room at 15°C with initial relative humidity 40%. The volume of the room is 50 $m^3$ . (a) How much water will evaporate ? (b) If the room temperature is increased by 5°C how much more water will evaporate ? The saturation vapour pressure of water at 15°C and 20°C are 1.6 kPa and 2.4 kPa respectively.

##### Solution :

(a) Rel. humidity = $\frac{VP}{SVP at 15^{\circ}C} \Rightarrow 0.4 = \frac{VP}{1.6\times 10^3} \Rightarrow VP = 0.4\times 1.6\times 10^3$$\\The evaporation occurs as along as the atmosphere does not become saturated.\\Net pressure change = 1.6 × 10^3 – 0.4 × 1.6 × 10^3 = (1.6 – 0.4 × 1.6)10^3 = 0.96 × 10^3$$\\$Net mass of water evaporated = $m \Rightarrow 0.96 × 10^3 × 50 = \frac{m}{18}\times 8.3\times 288$$\\$$\Rightarrow m = \frac{0.96\times 50\times 18\times 10^3}{8.3\times 288} = 361.45\approx 361 g$$\\(b) At 20°C SVP = 2.4 KPa, At 15°C SVP = 1.6 KPa\\Net pressure charge = (2.4 – 1.6) × 10^3 Pa = 0.8 × 10^3 Pa\\Mass of water evaporated = m' = 0.8 × 10^3 50 = \frac{m'}{18}\times 8.3\times 293$$\\$$\Rightarrow m' = \frac{0.8\times 50\times 18\times 10^3}{8.3\times 293} = 296.06 \approx$ 296 grams