Concept Of Physics Laws of Thermodynamics

H C Verma

Concept Of Physics

1.   A thermally insulated, closed copper vessel contains water at $15°C$. When the vessel is shaken vigorously for $15$ minutes, the temperature rises to $17°C$. The mass of the vessel is $100 g$ and that of the water is $200 g$. The specific heat capacities of copper and water are $\frac{420J}{kg-K}$ and $\frac{4200J}{kg-k}$ respectively. Neglect any thermal expansion, $(a)$ How much heat is transferred to the liquid—vessel system ? $(b)$ How much work has been done on this system ? $(c)$ How much is the increase in internal energy of the system ?

$t_1=15^o C t_2 = 17^0 C $ $\\$ $\Delta = t_2-t_1 = 17 -15 = 2^o C = 2+273 = 275 K$ $\\$ $ m_v =100g = 0.1 kg $ $\\$ $m_w= 200g = 0.2 kg$ $\\$ $cu_g =\frac{420J}{kg-k}$ $\\$ $w_g = \frac{4200J}{kg-k}$ $\\$ $a)$ Te heat transferred to the liquid vessel system is 0. The internal heat is shared in between the vessels and water. $\\$ $b)$ Work done on the system = Heat produced unit $\\$ $\Rightarrow dw=100 \times 10^{-3} \times 420 \times 2 +200 \times 10^{-3} \times 4200 \times 2 = 84+84 \times 20 = 84 \times 21 = 1764 J$ $\\$ $c) dQ = 0, dU = -dw = 1764$ [since $dw=-ve$ work done on the system]

2.   Figure $(26-E1)$ shows a paddle wheel coupled to a mass of $12 kg$ through fixed frictionless pulleys. The paddle is immersed in a liquid of heat capacity $\frac{4200J}{K}$ kept in an adiabatic container. Consider a time interval in which the $12 kg$ block falls slowly through $70 cm$. $(a)$ How much heat is given to the liquid ? $b)$ How much work is done on the liquid ? $(c)$ Calculate the rise in the temperature of the liquid neglecting the heat capacity of the container and the paddle.

$a)$ Heat is not given to the liquid. Instead the mechanical work done is converted to heat. So heat given to liquid is $z$ $\\$ $b)$ Work done on the liquid is the PE lost by the $12 kg$ mass = $mgh$ = $12 \times 10 \times 0.70 = 84 J$ $\\$ $c)$ Rise in temp at $\Delta t$ $\\$ We know , $84 = ms\Delta t$ $\\$ $\Rightarrow 84 = 1 \times 4200 \times \Delta t$ $(for 'm' = 1kg) \Rightarrow \Delta t =\frac{84}{4200} = 0.02 k$ $\\$

3.   A $100 kg$ block is started with a speed of$\frac{2.0m}{s}$ on a long, rough belt kept fixed in a horizontal position. The coefficient of kinetic friction between the block and the belt is $0.20$. $(a)$ Calculate the change in the internal energy of the block-belt system as the block comes to a stop on the belt, $(b)$ Consider the situation from a frame of reference moving at$\frac{2.0m}{s}$ along the initial velocity of the block. As seen from this frame, the block is gently put on a moving belt and in due time the block starts moving with the belt at $\frac{2.0m}{s}$. Calculate the increase in the kinetic energy of the block as it stops slipping past the belt, $(c)$ Find the work done in this frame by the external force holding the belt.

Mass ofblock = $100 kg$ $\\$ $u=\frac{2m}{s}, m=0.2$ $v=0$ $\\$ $dQ=du+dw$ $\\$ In this case $dQ=0$ $\\$ $\Rightarrow -du=dw \Rightarrow du = -\bigg(\frac{1}{2}mv^2 - \frac{1}{2}mu^2\bigg) = \frac{1}{2} \times 100 \times 2 \times 2 = 200J $ $\\$

4.   Calculate the change in internal energy of a gas kept in a rigid container when $100 J$ of heat is supplied to it

$Q = 100J$ $\\$ We know ,$ \Delta U= \Delta Q-\Delta W$ $\\$ Here since the container is rigid, $\Delta V=0,$ $\\$ Hence the $\Delta W = P\Delta V = 100J$ $\\$ So, $\Delta U = \Delta Q = 100J$ $\\$

5.   The pressure of a gas changes linearly with volume from $10 kPa, 200 cc$ to $50 kPa, 50 cc$. $(a)$ Calculate the work done by the gas. $(b)$ If no heat is supplied or extracted from the gas, what is the change in the internal energy of the gas ?

$P_1=10 kpa = 10 \times 10^3 pa$ $\\$ $P_2=50 \times 10^3 pa$ $\\$ $v_1 = 200cc$ $\\$ $v_2=50cc$ $\\$ $(i)$ Work done on the gas =$\frac{1}{2}(10+50)\times 10^3 \times (50-200) \times 10^{-6}=-4.5 J$ $\\$ $(ii)dQ = 0 \Rightarrow 0=du+dw\Rightarrow du=-dw=4.5J$ $\\$

6.   An ideal gas is taken from an initial state $i$ to a final state $f$ in such a way that the ratio of the pressure to the absolute temperature remains constant. What will be the work done by the gas ?

initial state $I$ $\\$ Final state $f$ $\\$ Given $\frac{P_1}{T_1}=\frac{P_2}{T_2}$ $\\$ where $P_1 \rightarrow$ Initial pressure ; $P_2 \rightarrow$ Final Pressure $\\$ $T_2, T_1 \rightarrow $Absolute temp. So, $\Delta V=0$ $\\$ Work done by gas= $P \Delta $V=0

7.   Figure $(26-E2)$ shows three paths through which a gas can be taken from the state $A$ to the state $B$. Calculate the work done by the gas in each of the three paths.

In path $ACB$ , $\\$ $W_{AC} +W_{BC} = 0 +pdv = 30 \times 10^3 (25-10) \times 10^{-6}=0.45 J$ $\\$ In path $ AB, W_{AB}=\frac{1}{2} \times (10+30) \times 1063 15 \times 10^{-6} = 0.30J$ $\\$ In path $ADB, W = W_{AD}+W_{DB}=10 \times 10^3(25-10) \times 10^{-6}+0=0.15J$

8.   When a system is taken through the process abc shown in figure $(26-E3)$, $80 J$ of heat is absorbed by the system and $30 J$ of work is done by it. If the system does $10 J$ of work during the process $adc$, how much heat flows into it during the process ?

$\Delta Q = \Delta U+ \Delta W $ $\\$ In abc, $\Delta Q = 80J$ $\\$ $ \Delta W=30 J$ $\\$ So, $\Delta U = (80-30)J = 50J $ $\\$ Now in $adc, \Delta W = 10J$ $\\$ So , $\Delta Q = 10+50 =60J [\therefore \Delta U = 50J]$ $\\$

9.   $50 cal$ of heat should be supplied to take a system from the state $A$ to the state $B$ through the path $ACB$ as shown in figure $(26-E4)$. Find the quantity of heat to be supplied to take it from $A$ to $B$ via $ADB$.

In path $ ACB$, $\\$ $dQ = 50.050 \times 4.2 = 210J$ $dW = W_{AC}+W_{CB} = 50 \times 10^3 \times 200 \times 10^{-6} = 10J$ $\\$ $dQ = dU + dW$ $\\$ $\Rightarrow dU = dQ-dW = 210 -10 = 200J$ $\\$ In path $ADB, dQ = ?$ $\\$ $dU= 200J$ (internal energy change between 2 points is always same) $\\$ $ dW=W_{AD}=W_{DB}=0+155 \times 10^3 \times 200 \times 10^{-6} = 31J$ $\\$ $dQ = dU+dW = 200+31=231 J = 55cal$ $\\$

10.   Calculate the heat absorbed by a system in going through the cyclic process shown in figure $(26-E5)$.

heat absorbed = work done = Area under the graph $\\$ In the given case heat absorbed = are of the circle =$\pi \times 10^4 \times 10^{-6} \times 10^3 = 3.14 \times 10 = 31.4 J$ $\\$

11.   A gas is taken through a cyclic process $ABC$ $A$ as shown in figure $(26-E6)$. If $2.4 cal$ of heat is given in the process, what is the value of $J$ ?

Answer

11   None

$dQ = 2.4\ cal = 2.4\ J\ Joules$ $\\$ $dw = W_{AB} + W_{BC} + W_{AC}$ $\\$ $= 0 + \Big(\dfrac{1}{2}\Big) \times (100 + 200) \times 10^3 200 \times 10^{–6} – 100 \times 103 \times 200 \times 10{–6}$ $\\$ $= \Big(\dfrac{1}{2}\Big) \times 300 \times 10^3 200 \times 10^{–6} – 20 = 30 – 20 = 10 joules.$ $\\$ $du = 0$ (in a cyclic process) $\\$ $dQ = dU + dW \Rightarrow 2.4\ J = 10$ $\\$ $\Rightarrow J = \dfrac{10}{2.4} \approx 4.17\ J/Cal.$

12.   A substance is taken through the process $abc$ as shown in figure $(26-E7)$. If the internal energy of the substance increases by $5000 J$ and a heat of $2625 cal$ is given to the system, calculate the value of $J$.

Now, $\Delta{Q} = (2625 \times J)\ J$ $\\$ $\Delta{U} = 5000\ J$ $\\$ From Graph $\Delta{W} = 200 \times 10^3 \times 0.03 = 6000\ J.$ $\\$ Now, $\Delta{Q} = \Delta{W} + \Delta{U}$ $\\$ $\Rightarrow 2625\ J = 6000 + 5000\ J$ $\\$ $J = \dfrac{11000}{2625} = 4.19\ J/Cal$

13.   A gas is taken along the path $AB$ as shown in figure $(26-E8)$. If $70 cal$ of heat is extracted from the gas in the process, calculate the change in the internal energy of the system.

$dQ = 70cal = (70 \times 4.2) J$ $\\$ $dW = (\frac{1}{2} \times (200+500) \times 10^3 \times 150 \times 10^{-6}$ $\\$ $=(\frac{1}{2}) \times 500 \times 150 \times 10^{-3}$ $\\$ $525 \times 10^{-1} = 52.5 J $ $\\$ $dU = ?$ $\\$ $dQ = du +dw$ $\\$ $\Rightarrow -294 = du +52.5$ $\\$ $\Rightarrow du = -294 -52.5 = -346.5 J $ $\\$

14.   The internal energy of a gas is given by $U = 1.5 pV$ . It expands from $100cm^3$ to $200 cm^3$ against a constant pressure of $1.0 \times 10^5 Pa$. Calculate the heat absorbed by the gas in the process.

$U = 1.5pV$ $\\$ $P= 1 \times 10^5 Pa$ $\\$ $dV = (200 - 100) - 52.5 = -346.5J $ $\\$ $dU = 1.5 \times 10^5 \times 10^-4 = 15$ $\\$ $dW = 10^5 \times 10^-4 = 10$ $\\$ $dQ = dU +dW = 10 +15 = 25J$ $\\$

15.   A gas is enclosed in a cylindrical vessel fitted with a frictionless piston. The gas is slowly heated for some time. During the process, $10 J$ of heat is supplied and the piston is found to move out $10 cm$. Find the increase in the internal energy of the gas. The area of cross-section of the cylinder = $4 cm^2$ and the atmospheric pressure =$100 kPa$.

$dQ = 10J$ $\\$ $dV = A \times 10cm^3 = 4 \times 10cm^3 = 40 \times 10^-6 cm^3 $ $\\$ $dw = Pdv = 100 \times 10^3 \times 40 \times 10^6 = 4cm^3$ $\\$ $du = ?$ $\\$ $ 10 =du +dw \Rightarrow 10 = du +4 \Rightarrow du = 6J$ $\\$

$dQ = 10J$ $\\$ $dV = A \times 10cm^3 = 4 \times 10cm^3 = 40 \times 10^-6 cm^3 $ $\\$ $dw = Pdv = 100 \times 10^3 \times 40 \times 10^6 = 4cm^3$ $\\$ $du = ?$ $\\$ $ 10 =du +dw \Rightarrow 10 = du +4 \Rightarrow du = 6J$ $\\$

16.   A gas is initially at a pressure of $100 kPa$ and its volume is $2.0m^3$.Its pressure is kept constant and the volume is changed from $2.0m^3$ to $2.5m^3$ . Its volume is now kept constant and the pressure is increased from $100 kPa$ to $200 kPa$. The gas is brought back to its initial state, the pressure varying linearly with its volume, $(a)$ Whether the heat is supplied to or extracted from the gas in the complete cycle ? $b)$ How much heat was supplied or extracted ?

$(a) = P_1 = 100KPa$ $\\$ $V_1 = 2m^3$ $\\$ $\Delta V_1 = 0.5 m^3$ $\\$ $\Delta P_1 = 100KPa$ $\\$ From the graph, we find that area under AC is greater than area under than AB. So we see that heat is extracted from the system $b)$ Amount of heat = Area under ABC $\\$ $=\frac{1}{2} = \frac{5}{10} = 10^5 = 25000J$ $\\$

17.   Consider the cyclic process $ABCA$, shown in figure (26-E9), performed on a sample of $2.0$ mole of an ideal gas. A total of $1200J$ of heat is withdrawn from the sample in the process. Find the work done by the gas during the part $BC$.

$ n= 2 mole$ $\\$ $dU =0$ $\\$ $dQ = -1200 J$ $\\$ $dU = 0 (During Cyclic Process)$ $\\$ $dQ = dU +dwc$ $\\$ $\Rightarrow -1200 = W_{AB}+ W_{BC}+W_{CA}$ $\\$ $\Rightarrow -1200 = nR\Delta T + W_{BC}+0$ $\\$ $\Rightarrow -1200 = 2\times 8.3 \times 200 + W_{BC}$ $\\$ $\Rightarrow W_{BC}=-400 \times 8.3 - 1200 = -4520 J$ $\\$

18.   Figure (26-E10) shows the variation in the internal energy $U$ with the volume $V$ of $2.0$ mole.of an ideal gas in a cyclic process $abcda$. The temperatures of the gas at $b$ and $c$ are $500 K$ and $300 K$ respectively. Calculate the heat absorbed by the gas during the process.

Given $n=2 moles$ $\\$ $ dv=0$ $\\$ in $ad$ and $bc$. $\\$ Hence $dW =dQ$ $\\$ $dW = dW_{ab}+dW_{cd}$ $\\$ $=nRT_1 L_n \frac{2V_o}{V_o}$ $\\$ $=nR \times 2.303 \times log 2(500-300)$ $\\$ $ = 2 \times 8.314 \times 2.303 \times 0.301 \times 200 = 2305.31 J$ $\\$

19.   Find the change in the internal energy of $2 kg$ of water as it is heated from $0°C to 4°C$. The specific heat capacity of water is 4200 JAg-K and its densities at $0°C$ and $4°C$ are $999-9\ kgm^3$ $\\$ and ${1000 kg}{m^3}$ $\\$ respectively. Atmospheric pressure = $10 " Pa$ $\\$.

Given $M =2kg$ $\\$ $2t = 4^oC$ $\\$ $Sw = \frac{4200J}{Kg-k}$ $\\$ $f_0 = \frac{999.9kg}{m^3}$ $\\$ $f_4 = \frac{1000kg}{m^3}$ $\\$ $P = 10^5Pa$ $\\$ Net internal energy = $dv$ $\\$ $dQ = DU +dw \Rightarrow ms\Delta Q\phi = dU +(v_0-v_4)$ $\\$ $\Rightarrow 2 \times 4200 \times 4 = dU + 10^5(m-m)$ $\\$ $\Rightarrow 33600 = dU + 10^5 \bigg(\frac{m}{v_0}-\frac{m}{v_4}\bigg) = dU + 10^5(0.0020002- 0.002) = dU + 10^50.0000002$ $\\$ $\Rightarrow 33600 = dU +0.02 \Rightarrow dU = (33600 - 0.02)J $ $\\$

20.   Calculate the increase in the internal energy of $10g$ of water when it is heated from $0°C to 100°C$ and converted into steam at $100kPa$. The density of steam = ${0.6km}{m^3}$. Specific heat capacity of water = $4200{J}{kg-°C}$ and the latent heat of vaporization of water = $2.25 \times \frac{10^6 J}{kg}$ $\\$.

Mass = $10g = 0.01 kg$ $\\$ $P = 10^5 Pa$ $\\$ $dQ = Q_{H_2 O} 0^o - 100^o + Q_{H_2 O} -steam $ $\\$ $= 0.01 \times 4200 \times 100 = 0.01 \times 2.5 \times 10^6 = 4200 + 25000 = 29200 $ $\\$ $dW = P \times \Delta V$ $\\$ $Delta = \frac{0.01}{0.6} - \frac{0.01}{1000} = 0.01699$ $\\$ $ dW = p\Delta V = 0.01699 \times 10^5 1699 J $ $\\$ $dQ = dW +dU or dU = dQ - dW = 29200 - 1699 = 27501 = 2.75 \times 10^4 J$ $\\$

21.   Figure (26-E11) shows a cylindrical tube of volume $V$ with adiabatic walls containing an ideal gas. The internal energy of this ideal gas is given by $1.5 nRT$. The tube is divided into two equal parts by a fixed diathermic wall. Initially, the pressure and the temperature are p_1,T_1 on the left and p_2, T_2 on the right. The system is left for sufficient time so that the temperature becomes equal on the two sides, $(a)$ How much work has been done by the gas on the left part ? $(b)$ Find the final pressures on the two sides, $(c)$ Find the final equilibrium temperature, $(d)$ How much heat has flown from the gas on the right to the gas on the left?

$a)$ Since the wall can not be moved thus $dU$ = $0$ and $dQ =0$ $\\$ Hence $dW = 0$ $\\$ $b$ Let final pressure in $LHS$ = $P_1$ $\\$ In $RHS = P_2$ $\\$ ($\therefore$ no. of moles remain constant ) $\\$ $\frac{P_1V}{2RT_1} = \frac{P_1V}{2RT}$ $\\$ $\Rightarrow P_1 = \frac{P_1V}{T_1} = \frac{P_1(P_1+P_2)T_1T_2}{\lambda}$ $\\$ As, $T = \frac{P_1(P_1+P_2)T_1T_2}{\lambda}$ $\\$ Similarly $P_2 = \frac{P_2T_1(P_1+P_2)}{\lambda}$ $\\$ $c)$ Let $T_2 > t_1$ be the common temp $\\$ Initially $\frac{p_1V}{2} = n_1$ $tr_1$ $\Rightarrow n_1 = \frac{P_1 V}{2RT _1}$ $\\$ $n_2 = \frac{P_2V}{2RT_2}$ Hence $dQ =0, dW =0$,$\\$ Hence $dU = 0$ $\\$ Incase $(LHS)$ $\Delta u_1 = 1.5n_1 R(T-T_1) But \Delta u_1 - u_2 = 0$ $\\$ $\Delta u_2 = 1.5n_2 R(T_2 - T) $ $\\$ $\Rightarrow 1.5n _1 R(T-T_1) =1.5 n_2 (T_2-T)$ $\\$ $\Rightarrow n_2 T-n_1 T_1 = n_2 T_2 - n_2 T\Rightarrow T(n_1 +n_2) = n_1T_1 +n_2T_2$ $\\$ $\Rightarrow T = \frac{n_1T_1+n_2T_2}{n_1+n_2}$ $\\$ $=\frac{\frac{P_1 V}{2RT_1}\times T_1 + \frac{P_2V}{2RT_2}\times T_2}{\frac{P_1 V}{2RT_1} +\frac{P_2V}{2RT_2} } = \frac{\frac{P_1+P_2}{P_1T_2+P_2T_1}}{T1T2} $ $\\$ $\frac{(P_1+P_2)T_1T_2}{P_1T_2+P_2T_1} = \frac{(P_1+P_2)T_1T_2}{\lambda} as p_1 T_2 = P_2 T_1 = \lambda$ $\\$ $d)$ For $RHS$ $dQ$ =$dU$(As $dW = 0$) $\\$ =$1.5 n_2 R(T_2-t)$ $\\$ $=\frac{1.5P_2V}{2RT_2} R \bigg[\frac{T_2-(P_1-P_2)T_1T_2}{P_1T_2-P_2T_1}\bigg] = \frac{1.5P_2V}{2T_2} \bigg(\frac{P_1t2 ^2 - P_1T_1T_2}{\lambda}\bigg) $ $\\$ $ = \frac{1.5P_2V}{2T_2} \times \frac{T_2P_1(T_2-Y_1)}{\lambda} = \frac{3P_1P_2(t_2-T_1)V}{4\lambda}$ $\\$

22.   An adiabatic vessel of total volume $V$ is divided into two equal parts by a conducting separator. The separator is fixed in this position. The part on the left contains one mole of an ideal gas $(U = 1.5 nRT)$ and the part on the right contains two moles of the same gas. Initially, the pressure on each side is $p$. The system is left for sufficient time so that a steady state is reached. Find $(a)$ the work done by the gas in the left part during the process, $(b)$ the temperature on the two sides in the beginning, $(c)$ the final common temperature reached by the gases, $(d)$ the heat given to the gas in the right part and $(e)$ the increase in the internal energy of the gas in the left part.

$a$ As the conducting wall is fixed the work done by the gas on the left part during the process is Zero $\\$ $b)$ For left side $\\$ Pressure = $P$ Volume = $V$ $\\$ For right side Let initial temperature =$T_2$ $\\$ No. of moles= $n$ (1mole) $\\$ Let intial temperature = $T_1$ $\\$ $\frac{PV}{2} = nRT_1$ $\\$ $\frac{PV}{2} = n_2RT_2 $ $\\$ $\Rightarrow \frac{PV}{2} =(1)RT_1$ $\\$ $\Rightarrow T_2 = \frac{PV}{2n_2R} \times 1$ $\\$ $\Rightarrow T_1 = \frac{PV}{2(moles)R}$ $\\$ $\Rightarrow T_2 = \frac{PV}{4(moles)R}$ $\\$ $c)$ Let the final temperature = $T$ $\\$ Final temperature = $R$ $\\$ No .of. mole = $1$ mole + $2$ moles = $3$ moles $\\$ $\therefore = PV = nRT \Rightarrow T = \frac{PV}{nR} = \frac{PV}{3(mole)R}$ $\\$ $d)$ For $RHS$ $dQ$ = $dU$ [as, $dW$ = 0]$\\$ = $1.5$ $n_2 R(T-T_2) = 1.5 \times 2 \times R \times \bigg[\frac{PV}{3(mole)R} - \frac{PV}{4(mole)R}\bigg]$ $\\$ $ = 1.5 \times 2 \times R \times \frac{4PV-3PV}{4 \times 3 (mole)} = \frac{3 \times R \times PV}{3 \times 4 \times R} = \frac{PV}{4}$ $\\$ $e)$ As, $dQ$ = $-dU$ $\\$ $\Rightarrow dU = -dQ = \frac{-PV}{4}$ $\\$