 # Specific Heat Capacities of Gases

## Concept Of Physics

### H C Verma

1   1. A vessel containing one mole of a monatomic ideal gas (molecular weight = 20 g/mol) is moving on a floor at a speed of 50 m/s. The vessel is stopped suddenly. Assuming that the mechanical energy lost has gone into the internal energy of the gas, find the rise in its temperature.

##### Solution :

N= 1 mole, W= 20g/mol, V=50m/s $\\$ K.E of the vessel= internal energy of the gas $\\$ $=(1/2)mv^2=(1/2)\times20\times10^{-3}\times50\times50=25J$ $\\$ 25=n$\frac{3}{2}r({\Delta T})\Rightarrow25=1\times\frac{3}{2}\times8.31\times\Delta T\Rightarrow\Delta T=\frac{50}{3\times8.3}=2k$ $\\$

2   2. 5 g of a gas is contained in a rigid container and is heated from 15°C to 25°C. Specific heat capacity of the gas at constant volume is 0.172 cal/g-°C and the mechanical equivalent of heat is 4.2 J/cal. Calculate the change in the internal energy of the gas.

##### Solution :

$\gamma =1.4$, w or piston=50kg, A of piston = 100cm$^2$ $\\$ Po= 100 kpa, g =10$m/s^2$ x=20cm $\\$ dw=pdv=$(\frac{mg}{A}+Po)Adx=(\frac{50\times 10}{100\times10^{-4}} + 10^5)100\times 10^{-4}\times 20\times 10^{-2}=1.5\times 10^5\times20\times10^{-4}=300J.$ $\\$ nRdt=300$\Rightarrow dT=\frac{300}{nR}$ $\\$ dQ=nCpdT=$nCp\times \frac{300}{nR}=\frac{n_\gamma R300}{(\gamma-1)nR}=\frac{300\times14}{0.4}=1500J.$ $\\$

3   3. Figure (27-E1) shows a cylindrical container containing oxygen (y - 1'4) and closed by a 50 kg friction less piston. The area of cross-section is 100 cm atmospheric pressure is 100 kPa and g is 10 m/s. The cylinder is slowly heated for some time. Find the amount of heat supplied to the gas if the piston moves out through a distance of 20 cm.

##### Solution :

$\gamma =1.4$, w or piston=50kg, A of piston = 100cm$^2$ $\\$ Po= 100 kpa, g =10$m/s^2$ x=20cm $\\$ dw=pdv=$(\frac{mg}{A}+Po)Adx=(\frac{50\times 10}{100\times10^{-4}} + 10^5)100\times 10^{-4}\times 20\times 10^{-2}=1.5\times 10^5\times20\times10^{-4}=300J.$ $\\$ nRdt=300$\Rightarrow dT=\frac{300}{nR}$ $\\$ dQ=nCpdT=$nCp\times \frac{300}{nR}=\frac{n_\gamma R300}{(\gamma-1)nR}=\frac{300\times14}{0.4}=1500J.$ $\\$

4   4. The specific heat capacities of hydrogen at constant volume and at constant pressure are 2.4 cal/g-°C and 3.4cal/g-°C respectively. The molecular weight of hydrogen is 2 g/mol and the gas constant R = 8.3 x 10 ' erg/mol-°C. Calculate the value of J.

##### Solution :

$C_vH_2=2.4 Cal/g^{\circ}C, \hspace{0.5cm}C_pH^2=3.4Cal/g^{\circ}C$ $\\$ M=2g/Mol, $R=8.3\times10^7 erg/mol-^{\circ}C$ $\\$ We know $C_p-C_v=1Cal/g^{\circ} C$ $\\$ So difference of molar specific heats $\\$ =$C_p\times M-C_v\times M=1\times2=2Cal/g^{\circ}C$ $\\$ Now $2\times J=R\Rightarrow2\times J=8.3\times10^7erg/mol-^{\circ} C \Rightarrow J=4.15\times10^7 erg/cal$ $\\$

5   5. The ratio of the molar heat capacities of an ideal gas is C$_p,/C_v$= 7/6. Calculate the change in internal energy of l1.0 mole of the gas when its temperature is raised by 78 50 K (a) keeping the pressure constant, (b) keeping the volume constant and (c) adiabatically.

##### Solution :

$\frac{C_p}{C_v}=7.6,n=1 mole, \Delta T=50K$ $\\$ (a) Keeping the pressure constant , dQ=du+dw. $\\$ $\Delta T=50K, \gamma =7/6,m=1mole.$ $\\$ $dQ=du+dw\Rightarrow C_vdT=du+RdT\Rightarrow du=nC_pdT-RdT$ $\\$ =$1\times\frac{R\gamma}{\gamma-1}\times dT-RdT=\frac{R\times\frac{7}{6}}{\frac{7}{6}-1}dT-RdT$ $\\$ =DT-RdT=7RdT-RdT=6RdT=$6\times8.3\times50=2490J$ $\\$ (b) Kipping volume constant, dv=$nC_vdT$ $\\$ =$1\times\frac{R}{\gamma-1}\times dt=\frac{1\times8.3}{\frac{7}{6}-1}\times 50$ $\\$ =$8.3\times 50\times 6=2490J$ $\\$ (c) Adiabetically dQ=0,du=-dw $\\$ =[$\frac{n\times R}{\gamma-1}(T_1-T_2)]=\frac{1\times8.3}{\frac{7}{6}-1}(T_2-T_1)=8.3\times50\times6=2490J$ $\\$

6   6. A sample of air weighing 1.18 g occupies I.0 $\times$$10^3cm 3 when kept at 300 K and l.0\times 10^5 Pa. When 2.0 cal of heat is added to it at constant volume, its temperature increases by 1°C. Calculate the amount of heat needed to increase the temperature of air by 1°C at constant pressure if the mechanical equivalent of heat is 4.2 x 10 ' erg/cal. Assume that air behaves as an ideal gas. ##### Solution : m=1.8g, V=1\times 10^3 cm^3=1L T=300k, P=10^5 Pa \\ PV=nRT or n=\frac{PV}{RT}=\frac{1}{8.2\times10^{-2}\times3\times10^2}=\frac{1}{8.2\times3}=\frac{1}{24.6} \\ Now, C=\frac{1}{n}\times\frac{Q}{dt}=24.6\times2=49.2 \\ C_p=R+C_v=1.987+49.2=51.87 \\ Q=nC_pdT=\frac{1}{24.6}\times51.187\times1=2.08 Cal \\ 7 7. An ideal gas expands from 100 cm 1 to 200 cm at a constant pressure of 2.0 x 10 5 Pa when 50 J of heat is supplied to it. Calculate (a) the change in internal energy of the gas, (b) the number of moles in the gas if the initial temperature is 300 K, (c) the molar heat capacity Cp at constant pressure and (d) the molar heat capacity Cu at constant volume. ##### Solution : V_1=100cm^3, v2=200cm^3,P=2\times10^5Pa,\Delta Q=50J \\ (a) \Delta Q=du+dw\Rightarrow50=du+20\times10^5(200-100\times10^{-6})\Rightarrow 50=du+20\Rightarrow du=30j \\ (b) 30=n\times \frac{3}{2}\times8.3\times300[U=\frac{3}{2}nRT for monoatomic] \\ \Rightarrow n =\frac{2}{3\times 83}=\frac{2}{249}=0.08 \\ (c)du=nC_vdT\Rightarrow C_v=dndTu=\frac{30}{0.008\times300}=12.5 \\ C_p=C_v+R=12.5+8.3=20.3$$\\$ (d)$C_v=12.5$ (Proved above) $\\$

8   8. An amount Q of heat is added to a monoatomic ideal gas in a process in which the gas performs a work Q/2 on its surrounding. Find the molar heat capacity for the process.

##### Solution :

Q=Amt of heat given $\\$ Work done=$\frac{Q}{2}, \Delta Q=W+\Delta U$ $\\$ for monoatomic gas $\Rightarrow\Delta U=Q-\frac{Q}{2}=\frac{Q}{2}$ $\\$ $V=n\frac{3}{2}RT=\frac{Q}{2}=nT\times\frac{3}{2}R=3R\times nT$ $\\$ Again Q=nCpdT Where C_p>Molar heat capacity at const. pressure $\\$ $3RnT=ndTC_p\Rightarrow C_p=3R$ $\\$

9   9. An ideal gas is taken through a process in which the pressure and the volume are changed according to the equation p = kV. Show that the molar heat capacity of the gas for the process is given by C = Cv + R/2 •

##### Solution :

P=kV$\Rightarrow \frac{nRT}{V}=KV\Rightarrow RT=KV^2\Rightarrow R \Delta T=2KV \Delta U \Rightarrow \frac{R\Delta T}{2KV}=dv$ $\\$ dQ=du+dw$\Rightarrow mcdT=C_vdT+pdv\Rightarrow msdT=C_vdT+\frac{PRdF}{2KV}$ $\\$ $\Rightarrow ms=C_v+\frac{RKV}{2KV}\Rightarrow C_p+\frac{R}{2}$ $\\$

10   10. An ideal gas (Cp/Cv= y) is taken through a process in which the pressure and the volume vary as p = aV$^b$. Find the value of b for which the specific heat capacity in the process is zero.

##### Solution :

$\frac{C_p}{C_v}=\gamma$ $C_p-C_v=R$, $C_v=\frac{r}{\gamma-1}$ $C_p=\frac{\gamma R}{\gamma-1}$ $\\$ PdV=$\frac{1}{b+1}(Rdt)$ $\\$ $\Rightarrow 0=C_vdT+\frac{1}{b+1}(Rdt)\Rightarrow \frac{1}{b+1}=\frac{-C_v}{R}$ $\\$ $\Rightarrow b+1=\frac{-R}{C_V}=\frac{-(C_p-C_v)}{C_v}=-\gamma+1\Rightarrow b=-\gamma$ $\\$

11   11. Two ideal gases have the same value of Cp/Cv =$\gamma$. What will be the value of this ratio for a mixture of the two gases in the ratio 1:2?

Considering two gases in Gas(1) we have, $\\$ $\gamma Cp1(Sp. Heat Const. 'P'),Cv1(Sp. Heat at const. 'V') n_1(No. of moles)$ $\\$ $\frac{Cp_1}{Cv_1}=\gamma \& Cp1-Cv1=R$ $\\$ $\Rightarrow Cv_1-Cv_1=R\Rightarrow Cv_1(\gamma-1)=R$ $\\$ $\Rightarrow Cv_1=\frac{R}{\gamma-1} \& Cp_1=\frac{\gamma R}{\gamma -1}$ $\\$ In gas(2) we have , $\gamma , Cp_2(Sp.Heat at const'P'),Cv_2,n2$ $\\$ $\frac{Cp_2}{Cv_2}=\gamma \& Cp_2-Cv_2=R\Rightarrow \gamma Cv_2-Cv_2=R \Rightarrow Cv_2(\gamma-1)=R \Rightarrow Cv_2=\frac{R}{\gamma-1} \& Cp_2=\frac{\gamma R}{\gamma-1} $$\\ Given n_1:n_2=1:2 \\ dU_1=nCv_1dT \& dU_2=2nCv_2dT=3nCvdT \\ \Rightarrow Cv=\frac{CV_1+2Cv_2}{3}=\frac{frac\{R}{\gamma-1}+\frac{2R}{\gamma-1}}{3}=\frac{3R}{3(\gamma-1)}=\frac{R}{\gamma-1} ...(1) \\ \&Cp=\gammaCv=\frac{\gammar}{\gamma-1} ....(2) \\ So,\frac{Cp}{Cv}=\gamma [from(1) & (2)\ \\ 12 12. A mixture contains 1 mole of helium (Cp = 2*5 R, Cv =1.5 R) and 1 mole of hydrogen (Cp = 3.5 R, Cv = 2.5 R). Calculate the values of Cp, Cu and \gamma for the mixture. ##### Solution : Cp'=2.5R Cp''=3.5R \\ Cv'=1.5R and Cv''=2.5R \\ n_1=n_2= 1 mol \hspace{0.3cm}$$(n_1+n_2)C_vdT+n_2C_vdT$ $\\$ $\Rightarrow Cv=\frac{n_1Cv'+n_2Cv"}{n_1+n_1}=\frac{1.5R+2.5R}{2}2R$ $\\$ Cp=Cv+R=2R+R=3R $\\$ $\gamma=\frac{Cp}{Cv}=\frac{3R}{2R}=1.5$ $\\$

13   13. Half mole of an ideal gas ($\gamma$ = 5/3) is taken through the cycle abcda. Take R = 25/3 J/mol-K (a) Find the temperature of the gas in the states a, b, c and d. b) Find the amount of heat supplied in the processes ab and be. (c) Find the amount of heat liberated in the processes cd and da.

##### Solution : n=1/2 mole R=25/3 J/mol-k, $\gamma=\frac{5}{3}$ $\\$ (a) Temp at A=$T_a,P_aV_a=nRT_a$ $\\$ $\Rightarrow T_a=\frac{P_aV_a}{nR}=\frac{5000\times10^{-6}\times100\times10^3}{\frac{1}{2}\times\frac{25}{3}}=120k.$ $\\$ Similarly temperatures at point b=240k at C it is 480k and at D it is 240k. $\\$ (b)For ab process, $\\$ $dQ=nC_pdT$ [Since ab is isobaric] $\\$ =$\frac{1}{2}\times\frac{R_gamma}{\gamma-1}(T_b-T_a)=\frac{1}{2}\times\frac{\frac{25}{5}\times\frac{5}{3}}{\frac{5}{3}-1}\times(240-120)=\frac{1}{2}\times\frac{125}{9}\times \frac{3}{2}\times 120=1250J$ $\\$ For bc, dQ=dU+dW [dq=0, Isochorie process] $\\$ $\Rightarrow dQ=du=nC_vdT=\frac{nR}{\gamma-1}(T_c-T_c)=\frac{1}{2}\times \frac{\frac{25}{3}}{\frac{5}{3}-1}(240)=\frac{1}{2}\times{25}{3}\times\frac{3}{2}\times 240=1500 J$ $\\$ (c) Heat liberated in cd=-nCpdT $\\$ =$\frac{-1}{2}\times\frac{nR}{\gamma-1}(T_d-T_c)=\frac{-1}{2}\times \frac{125}{3}\times\frac{3}{2}\times240=2500J$ $\\$ Heat liberated in da=-nCvdT =$\frac{-1}{2}\times\frac{R}{\gamma-1}(T_a-T_d)=\frac{-1}{2}\times \frac{25}{2}\times(120-240)=750J$ $\\$

14   14. An ideal gas ($\gamma$ - 1.67) is taken through the process abc shown in figure (27-E3). The temperature at the point a is 300 K. Calculate (a) the temperatures at b and e, fa) the work done in the process, (c) the amount of heat supplied in the path ab and in the path be and (d) the change in the internal energy of the gas in the process.

##### Solution : (a) For a,b'V' is constant $\\$ So,$\frac{P_1}{T_1}=\frac{P_2}{T_2}\Rightarrow \frac{100}{300}=\frac{200}{T_2}\Rightarrow T_2=\frac{200\times300}{100}=600k$ $\\$ For, b,c,'P' is constant $\\$ So, \$\frac{V_1}{T_1}=\frac{V_2}{T_2}\Rightarrow\frac{100}{600}=\frac{150}{T_2}\Rightarrow T_2=\frac{600\times150}{100}=900K$ $\\$ (b) Work done=Area enclosed under the graph 50cc$\times200kpa=50\times10^{-6}\times 200\times10^3J=10J$ $\\$ (c)'Q' Supplied =nCvdT $\\$ Now,n=$\frac{PV}{RT}$considering at pt 'b' $\\$ Cv=$\frac{R}{\gamma-1}dT=300a,b.$ $\\$ Q$_bc=\frac{PV}{RT}\times \frac{R}{\gamma-1}dT=\frac{200\times10^3\times100\times10^{-6}}{600\times0.67}\times 300=14.925$ ($\therefore$$\gamma=1.67) \\ Q supplied to be nCpdT [\therefore C_p=\frac{\gamma R}{\gamma-1}] \\ =\frac{PV}{RT}\times\frac{\gamma R}{\gamma-1}dT=\frac{200\times10^3\times150\times10^{-6}}{8.3\times900}\times\frac{1.67\times8.3}{0.67}\times300=24.925 \\ (d)Q=\Delta U +w$$\\$ Now,$\Delta U=Q-w$=Heat supplied-Work done=(24.925+14.925)-1=29.850 $\\$

15   15. In Joly's differential steam calorimeter, 3 g of an ideal gas is contained in a rigid closed sphere at 20°C. The sphere is heated by steam at 100°C and it is found that an extra 0.095 g of steam has condensed into water as the temperature of the gas becomes constant. Calculate the specific heat capacity of the gas in J/g-K. The latent heat of vaporization of water = 540 cal/g.

##### Solution :

In jolly's differential steam calorimeter $\\$ $C_v=\frac{m_2L}{m1(\theta_2-\theta_1)}$ $\\$ m2=Mass of steam condensed= 0.095g, L=540Cal/g=540x2.4J/g $\\$ m1=mass of gas present =3g, $\theta_1=20^{\circ}C, \hspace{0.25cm} \theta_2=100^{\circ} C$ $\\$ $\Rightarrow C_v=\frac{0.095\times540\times4.2}{3(100-20)}=0.89=0.9 J/g-K$ $\\$ $\gamma=1.5$ $\\$

16   16. The volume of an ideal gas ($\gamma$ = 1.5) is changed adiabatically from 4.00 litres to 3.00 litres. Find the ratio of (a) the final pressure to the initial pressure and fa) the final temperature to the initial temperature.

##### Solution :

$\gamma=1.5$ $\\$ Since it is an adiabatic process, $PV^{\gamma}$=const. $\\$ (a) $P_1V_1^{\gamma}=P_2V_2^{\gamma} \hspace{0.25cm}Given V_1=4L_1,V_2=3L, \frac{P_2}{P_1}=?$ $\\$ $\Rightarrow \frac{P_2}{P_1}=(\frac{V_1}{V_2})^{\gamma}=(\frac{4}{3})^{1.5}=1.5396=1.54$ $\\$ (b) $TV^{-1}$=Const. $\\$ $T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1} \Rightarrow \frac{T_2}{T_1}=(\frac{V_1}{V_2})^{\gamma-1}=(\frac{4}{3})^{0.5}=1.154$ $\\$

17   17. An ideal gas at pressure 2.5 x 10.0 Pa and temperature 300 K occupies 100 cc. It is adiabatically compressed to half its original volume. Calculate (a) the final pressure, a) the final temperature and (c). the work done by the gas in the process. Take $\gamma$= 1.5.

##### Solution :

$P_1=2.5\times10^5Pa,\hspace{0.3cm} V_1=100cc, \hspace{0.2cm} T_1=300cc$ $\\$ (a)$P_1V_1^{\gamma}=P_2V_2^{\gamma}$ $\\$ $\Rightarrow2.5\times 10^5\times V^{1.5}=(\frac{V}{2})^1.5\times P_2$ $\\$ $\Rightarrow P_2=2^{1.5}\times 2.5\times 10^5=7.07\times10^5\approx 7.1\times10^5$ $\\$ (b)$T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}\Rightarrow 300\times(100)^{1.5-1}=T_2\times (50)^{1.5-1}$ $\\$ $\Rightarrow T_2=\frac{3000}{7.07}=423.32k=424k$ $\\$ (c) Work done by the gas in the process $\\$ W=$\frac{mR}{\gamma-1}[T_2-T_1]=\frac{P_1V_1}{T(\gamma-1)}[T_2-T_1]$ $\\$ $=\frac{2.5\times10^5\times100\times10^{-6}}{300(1,5-1)}[424-300]=\frac{2.5\times10}{300\times0.5}\times124=20.72\approx21J$ $\\$

18   18. Air (y$\gamma$- 1.4) is pumped at 2 atm pressure in a motor tyre at 20°C. If the tyre suddenly bursts, what would be the temperature of the air coming out of the tyre. Neglect any mixing with the atmospheric air.

##### Solution :

$\gamma=1.4,\hspace{0.2cm}T_1=20^{\circ}C=293k,\hspace{0.2cm}P_1=2 atm \hspace{0.1cm} p_2=1atm$ $\\$ We know for adiabatic process $\\$ $P_1^{1-\gamma}\times T_1^{\gamma}=P_2^{1-\gamma}\times T_2^{\gamma} or (2)^{1-1.4}\times(293)^{1.4}=(1)^{1-1.4}\times T_2^{1.4}$ $\\$ $\Rightarrow (2)^{0.4}\times(293)^{1.4}=T_2^{1.4}\Rightarrow 2153.78=T_2^{1.4}\Rightarrow T_2=(2153.78)^{1/1.4}=240.3k,$ $\\$

19   19. A gas is enclosed in a cylindrical can fitted with a piston. The walls of the can and the piston are adiabatic. The initial pressure, volume and temperature of the gas are 100 kPa, 400 $cm^3$ and 300 K respectively. The ratio of the specific heat capacities of the gas is Cp/Cv = l.5. Find the pressure and the temperature of the gas if it is (a) suddenly compressed fa) slowly compressed to 100 cm $^3$

##### Solution :

Given$\frac{Cp}{Cv}=\gamma \hspace{0.25cm}P_0(Inital Pressure),\hspace{0.25cm} V_0(Initial Volume)$ $\\$ (a) (i) Isothermal compression $P_1V_1=P_2V_2$ or $P_0V_0=\frac{P_2v_0}{2}\Rightarrow P_2=2P_0$ $\\$ (ii) Adiabatic Compression $P_1V_1^{\gamma}=P_2V_2^{\gamma} or 2p_0(\frac{V_0}{2})^\gamma=p1(\frac{V_0}{4})^\gamma$ $\\$ $\Rightarrow P'=\frac{V_0^\gamma}{2^\gamma}\times2P_0\times\frac{4^\gamma}{V_0^\gamma}=2^\gamma\times 2 P_0\Rightarrow P_02^{\gamma+1}$ $\\$ (b) (i)Adiabatic compression $P_1V_1^\gamma=P_2V_2^\gamma or P_0V_0^\gamma=P'(\frac{V_0}{2})^\gamma=P'=P_02^\gamma$ $\\$ (ii)Isothermal compression $P_1V_1=P_2V_2$ or $2^\gamma P_0\times\frac{V_0}{2}=P_2\times\frac{V_0}{4}\Rightarrow P_2=P_02^{\gamma+1}$ $\\$

20   20. The initial pressure and volume of a given mass of a gas (Cp/Cv - $\gamma$) are $p_0$ and $V_0$. The gas can exchange heat with the surrounding, (a) It is slowly compressed to a volume $V_0$/2 and then suddenly compressed to $V_0$/4. Find the final pressure, fa) If the gas is suddenly compressed from the volume $V_0$ to $V_0$/2 and then slowly compressed to $V_0$/4, what will be the final pressure ?

##### Solution :

Given$\frac{Cp}{Cv}=\gamma \hspace{0.25cm}P_0(Inital Pressure),\hspace{0.25cm} V_0(Initial Volume)$ $\\$ (a) (i) Isothermal compression $P_1V_1=P_2V_2$ or $P_0V_0=\frac{P_2v_0}{2}\Rightarrow P_2=2P_0$ $\\$ (ii) Adiabatic Compression $P_1V_1^{\gamma}=P_2V_2^{\gamma} or 2p_0(\frac{V_0}{2})^\gamma=p1(\frac{V_0}{4})^\gamma$ $\\$ $\Rightarrow P'=\frac{V_0^\gamma}{2^\gamma}\times2P_0\times\frac{4^\gamma}{V_0^\gamma}=2^\gamma\times 2 P_0\Rightarrow P_02^{\gamma+1}$ $\\$ (b) (i)Adiabatic compression $P_1V_1^\gamma=P_2V_2^\gamma or P_0V_0^\gamma=P'(\frac{V_0}{2})^\gamma=P'=P_02^\gamma$ $\\$ (ii)Isothermal compression $P_1V_1=P_2V_2$ or $2^\gamma P_0\times\frac{V_0}{2}=P_2\times\frac{V_0}{4}\Rightarrow P_2=P_02^{\gamma+1}$ $\\$

21   21. Consider a given sample of an ideal gas (Cp/Cu = y) having initial pressure p0 and volume V0. (a) The gas is isothermal ly taken to a pressure p0/2 and from there adiabatically to a pressure p0/4. Find the final volume, fa) The gas is brought back to its initial state. It is adiabatically taken to a pressure $p_0/2$ and from there isothermally to a pressure $p_0/4$. Find the final volume

##### Solution :

Initial pressure =$P_0$ Inital volume=$V_0$ $\\$ $\gamma=\frac{Cp}{Cv}$ $\\$ (a)Isothermally to pressure $\frac{P_0}{2}$ $\\$ $P_0V_0=\frac{P_0}{2}V_1\Rightarrow V_1=2V_0$ $\\$ Adiabetically to pressure=$\frac{P_0}{4}$ $\\$ $\frac{P_0}{2}(V_1)^\gamma \Rightarrow\frac{P_0}{2}(2V_0)^\gamma=\frac{P_0}{4}(V_2)^\gamma$ $\\$ $\Rightarrow 2^{\gamma+1}V_0^\gamma=V_2^{\gamma}\Rightarrow V_2=2^{(\gamma+1)/\gamma}V_0$ $\\$ $\therefore$ Final Volume=$2^{(\gamma+1)\gamma}V_0$ $\\$ (b)Adiabetically to pressure $\frac{P_0}{2} to P_0$ $\\$ $P_0\times(2^{\gamma+1}V_0{^\gamma})=\frac{P_0}{2}\times(V')^\gamma$$\\$ Isothermal to pressure $\frac{P_0}{4}$ $\\$ $\frac{P_0}{2}\times2^{1/\gamma}V_0=\frac{P_0}{4}\times V''\Rightarrow V''=2^{(\gamma+1)\gamma}V_0=The Final Volume$ $\\$

22   22. A sample of an ideal gas ($\gamma$=l.5) is compressed adiabatically from a volume of 150 $cm^3$ to 50 $cm^3$. The initial pressure and the initial temperature are 150 KPa and 300 K. Find (a) the number of moles of the gas in the sample, b) the molar heat capacity at constant volume, (c) the final pressure and temperature, (d) the work done by the gas in the process and (e) the change in internal energy of the gas.

##### Solution :

PV=nRT $\\$ Given P=150KPa=150x$10^3Pa,$ V=$150cm^3=150 \times 10^{-6}m^3, T=300K$ $\\$ (a)n=$\frac{PV}{RT}=\frac{150\times10^3\times150\times10^{-6}}{8.3\times300}=9.036\times10^{-3}=0.009 moles$ $\\$ (b) $\frac{Cp}{Cv}=\gamma\Rightarrow \frac{\gamma R}{(\gamma-1)Cv}=\gamma \hspace{0.2cm} [\therefore Cp=\frac{\gamma R}{\gamma-1}]$ $\\$ $\Rightarrow Cv=\frac{R}{\gamma-1}=\frac{8.3}{1.5-1}=\frac{8.3}{0.5}=2R=16.6J/mole$ $\\$ (c) Given $P_1=150KPa=150\times10^3Pa,\hspace{0.2cm} P_2=?$ $\\$ $V_1=150cm^3=150\times10^{-6}m^3,\gamma=1.5$ $\\$ $V_2=50cm^3=50\times10^{-6}m^3,\hspace{0.25cm} T_1=300k, T2=?$ $\\$ Since the process is adiabatic, $P_1V_1^\gamma=P_2V_2^\gamma$ $\\$ $\Rightarrow 150\times 10^3(150\times10^{-6})^\gamma=P_2\times(50\times10^{-6})^\gamma$ $\\$ $\Rightarrow P_2=150\times10^3\times(\frac{150\times10^{-6}}{50\times10^-{6}})^{1.5} =150000\times3^{1.5}=779.422\times10^3Pa=780KPa$ $\\$ (d) $\Delta Q=W+\Delta U or W=-\Delta U$ [$\therefore\Delta U=0$, in adiabetic]$\\$ =-nCvdT=$-0.009\times 16.6\times(520-300)=-0.009\times16.6\times220=-32.8J\approx33J$ $\\$ (e) $\Delta U=nCvdT=0.009\times16.6\times220\approx33J$ $\\$

23   23. Three samples A, B and C of the same gas ($\gamma= 1.5$) have equal volumes and temperatures. The volume of each sample is doubled, the process being isothermal for A, adiabatic for B and isobaric for C. If the final pressures are equal for the three samples, find the ratio of the initial pressures.

##### Solution :

$V_A=V_B=V_C$ $\\$ For A, the process is isothermal $\\$ $P_AV_A=P_A'V_A'\Rightarrow P_A'=P_A\frac{V_A}{V_A'}=P_A\times \frac{1}{2}$ $\\$ For B, the process is adiabetic, $\\$ $P_A(V_B)^\gamma=P_A'(V_B)^\gamma=P_B'=P_B(\frac{V_B}{V_B'})^\gamma=P_B\times (\frac{1}{2})^{1.5} = \frac{P_B}{2^{1.5}}$ $\\$ For C, the process is isobaric $\\$ $\frac{V_C}{T_C}=\frac{V_C'}{T_C'}\Rightarrow \frac{V_C}{T_C}=\frac{2V_C'}{T_C'}\Rightarrow T_C'=\frac{2}{T_C}$ $\\$ Final pressures are equal. $\\$ =$\frac{p_A}{2}=\frac{P_B}{2^{1.5}}=P_C\Rightarrow P_A:P_B:P_C=2:2^{1.5}:1=2\sqrt{2}:1$ $\\$

24   24. Two samples A and B of the same gas have equal volumes and pressures. The gas in sample A is expanded isothermally to double its volume and the gas in B is expanded adiabatically to double its volume. If the work done by the gas is the same for the two cases, show that y satisfies the equation $1 - 2 ^{1-\gamma} = (y - 1) ln2$.

##### Solution :

$P_1$= Initial pressure $V_1$=Initial Volume $P_2$=Final Pressure $V_2$=Final Volume $\\$ Given, $V_2=2V_1$ Isothermal work done =nR$T_1$Ln$(\frac{V_2}{V_3})$ $\\$ Adiabetic work done=$\frac{P_1V_1-P_2V_2}{\gamma-1}$ $\\$ $\Rightarrow (\gamma-1)ln(\frac{V_2}{V_1})=\frac{nRt_1-nRt_2}{nRT_1}\Rightarrow(\gamma-1)ln2=\frac{T_1-T_1}{T_1}$ ..(i) [$\therefore V_2=2V_1$] $\\$ We know that $TV^{-1}$=const. in adiabetic process. $\\$ $T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}, \hspace{0.15cm}or\hspace{0.15cm} T_1(V_2)^{\gamma-1}=T_2\times(2)^{\gamma-1}\times(V_1)^{\gamma-1}$ $\\$ Or $T_1=2^{\gamma-1}\times T_2 or T_2=T_1^{1-\gamma}$ ..(ii) $\\$ From (i) and (ii) $\\$ $(\gamma-1)ln2=\frac{T_1-T_1\times2^{1-\gamma}}{T}\Rightarrow (\gamma-1)ln2=1-2^{1-\gamma}$ $\\$

25   25. 1 litre of an ideal gas ($\gamma$ = 1.5) at 300 K is suddenly compressed to half its original volume, (a) Find the ratio of the final pressure to the initial pressure. fa) If the original pressure is 100 KPa, find the work done by the gas in the process, (c) What is the change in internal energy ? (d) What is the final temperature ? (e) The gas is now cooled to 300 K keeping its pressure constant. Calculate the work done during the process. (0 The gas is now expanded isothermally to achieve its original volume of 1 litre. Calculate the work done by the gas. (g) Calculate the total work done in the cycle.

##### Solution :

$\gamma=1.5$ T=300K, V=1Lv=1/2l $\\$ The process is adiabatic as it is sudden, $\\$ $P_1V_1^{\gamma}=P_2V_2^{\gamma}\Rightarrow P_1(V_0)^{\gamma}=P_2(\frac{V_0}{2})^{\gamma}\Rightarrow P_2=P_1(\frac{1}{1/2})^{1.5}=P_1(2)^{1.5}\Rightarrow\frac{P_2}{P_1}=2^{1.5}=2\sqrt{2}$ $\\$ (b)$P_1$=100KPa=$10^5$ Pa W=$\frac{nR}{\gamma-1}[T_1-T_2]$ $\\$ $T_1V_1^{\gamma-1}=P_2V_2^{\gamma-1}\Rightarrow 300\times(1)^{1.5-1}=T_2=(0.5)^{1.5-1}\Rightarrow 300\times1=T_2\sqrt{0.5}$ $\\$ $T_2=300\times\sqrt{\frac{1}{0.5}}=300\sqrt{2}K$ $\\$ $P_1V_1=nRT_1\Rightarrow n=\frac{P_1V_1}{RT_1}=\frac{10^5\times10^3}{R\times300}=\frac{1}{3R}$ (V in $m^3$) $\\$ $w=\frac{nR}{\gamma-1}[T_1-T_2]=\frac{1R}{3R(1.5-1)}[300-300\sqrt{2}]=\frac{300}{3\times0.5}(1-\sqrt{2})=-82.8J\approx-83J$ $\\$ (c)Internal Energy $\\$ dQ=0,$\Rightarrow dU=-dW=-(-82.8J)=82.8J\approx82J$ $\\$ (d) The pressure is kept constant $\therefore$ The process is isobaric $\\$ Work done=nRdT=$\frac{1}{3R}\times R\times (300-300\sqrt{2})$ Final temperature= 300K $\\$ $-\frac{1}{3}\times300(0.414)=-41.4J.$ Initial Temp=$300\sqrt{2}$ $\\$ (f)Initial Volume$\Rightarrow \frac{V_1}{T_1}=\frac{V_1'}{T_1}=V_1'=\frac{V_1}{T_1}\times T_1' = \frac{1}{2\times300\times\sqrt{2}}\times 300=\frac{1}{2\sqrt{2}}L.$ $\\$ Final Volume=1L $\\$ Work done in isothermal=$nRTln\frac{V_2}{V_1}$ $\\$ =$\frac{1}{3R}\times R\times300ln(\frac{1}{1/2\sqrt{2}})=100\times ln(2\sqrt{2})=100\times 1.039=1.03 (g)Net work done=W$_A$+W$_B$+W$_C$=-82-41.4+103=-20.4J$ $\\$

26   26. Figure (27-E4) shows a cylindrical tube with adiabatic walls and fitted with an adiabatic separator. The separator can be slid into the tube by an external mechanism. An ideal gas ($\gamma$= 1.5) is injected in the two sides at equal pressures and temperatures. The separator remains in equilibrium at the middle. It is now slid to a position where it divides the tube in the ratio 1 : 3. Find the ratio of the temperatures in the two parts of the vessel

##### Solution : Given $\gamma=1.5$ $\\$ We know that for adiabatic process T$V^{-1}=const.$ $\\$ So $T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$ ..(eq) $\\$ As, it is an adiabatic process and all the other conditions are same, the above equation can be applied $\\$ So, $T_1\times(\frac{3V}{4})^{1.5-1}=T_2\times(\frac{V}{4})^{1.5-1} \Rightarrow T_1\times(\frac{3V}{4})^{0.5}=T_2\times(\frac{V}{4})^{0.5}$ $\\$ $\Rightarrow \frac{T_1}{T_2}=(\frac{V}{4})^{0.5} \times (\frac{4}{3V})^{0.5}=\frac{1}{\sqrt{3}} \hspace{0.35cm} So, T_1:T_2=1:\sqrt{3}$ $\\$

27   27. Figure (27-E5) shows two rigid vessels A and B, each of volume 200 $cm^3$ containing an ideal gas (Cv = 12.5 J/mol-K). The vessels are connected to a manometer tube containing mercury. The pressure in both the vessels is 75 cm of mercury and the temperature is 300 K. (a) Find the number of moles of the gas in each vessel, (b) 5.0 J of heat is supplied to the gas in the vessel .4 and 10 J to the gas in the vessel B. Assuming no appreciable transfer of heat from A to B calculate the difference in the heights of mercury in the two sides of the manometer. Gas constant R = 8.3 J/mol-K.

##### Solution : mHe=0.1g $\gamma$=1.67\mu$=4g/mol mHz=?$ $\\$ Since it is an adiabatic surrounding, $\\$ HedQ=nCvdT=$\frac{0.1}{4}\times\frac{R}{\gamma-1}\times dT=\frac{0.1}{4}\times\frac{R}{(1.67-1)}\times dT ..(i)$ $\\$ $H_2=nCvdT=\frac{m}{2}\times\frac{R}{\gamma-1}\times dT=\frac{m}{2}\times\frac{R}{1.4-1}\times dT$ [Where m is the rqd. Mass of $H_2$] $\\$ Since an equal amount of heat is given to both and $\Delta$ T is same in both, $\\$ Equating(i) and (ii), we get $\\$ $\frac{0.1}{4}\times\frac{R}{0.67}\times dT=\frac{m}{2}\times\frac{R}{0.4}\times dT \Rightarrow m=\frac{0.1}{2}\times\frac{0.4}{0.67}=0.0298=0.03g$ $\\$

28   28. Figure (27-E6) shows two vessels with adiabatic walls, one containing 0.1 g of helium ($\gamma$ = 1.67, M =4 g/mol) and the other containing some amount of hydrogen ($\gamma$= 1.4, M =2 g/mol). Initially, the temperatures of the two gases are equal. The gases are electrically heated for some time during which equal amounts of heat are given to the two gases. It is found that the temperatures rise through the same amount in the two vessels. Calculate the mass of hydrogen.

##### Solution : mHe=0.1g $\gamma$=1.67\mu$=4g/mol mHz=?$ $\\$ Since it is an adiabatic surrounding, $\\$ HedQ=nCvdT=$\frac{0.1}{4}\times\frac{R}{\gamma-1}\times dT=\frac{0.1}{4}\times\frac{R}{(1.67-1)}\times dT ..(i)$ $\\$ $H_2=nCvdT=\frac{m}{2}\times\frac{R}{\gamma-1}\times dT=\frac{m}{2}\times\frac{R}{1.4-1}\times dT$ [Where m is the rqd. Mass of $H_2$] $\\$ Since an equal amount of heat is given to both and $\Delta$ T is same in both, $\\$ Equating(i) and (ii), we get $\\$ $\frac{0.1}{4}\times\frac{R}{0.67}\times dT=\frac{m}{2}\times\frac{R}{0.4}\times dT \Rightarrow m=\frac{0.1}{2}\times\frac{0.4}{0.67}=0.0298=0.03g$ $\\$

29   29. Two vessels A and B of equal volume $V_0$ are connected by a narrow tube which can be closed by a valve. The vessels are fitted with pistons which can be moved to change the volumes. Initially, the valve is open and the vessels contain an ideal gas (Cp/Cv = $\gamma$) at atmospheric pressure p0 and atmospheric temperature T0 . The walls of the vessel A are diathermic and those of B are adiabatic. The valve is now closed and the pistons are slowly pulled out to increase the volumes of the vessels to double the original value, (a) Find the temperatures and pressures in the two vessels, fa) The valve is now opened for sufficient time so that the gases acquire a common temperature and pressure. Find the new values of the temperature and the pressure.

##### Solution : Initial pressure= $P_0$ $\\$ Initial Volume=$V_0$ $\\$ Initial Temperature=$T_0$ $\\$ $\frac{C_P}{C_V}=\gamma$ $\\$ (a) For the diathermic vessel the temperature inside remains constant $\\$ $P_1V_1-P_2V_2 \Rightarrow P_0V_0=P_2\times 2V_0 \Rightarrow P_2=\frac{P_0}{2}, \hspace{0.15cm}Temperature =T_0$ $\\$ For adiabatic vessel the temperature does not remain constant. The process is adiabatic. $\\$ $T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1} \Rightarrow T_0V_0^{\gamma-1}=T_2 \times (2V_0)^{\gamma-1} \Rightarrow T_2=T_0(\frac{V_0}{2V_0})^{\gamma-1}=T_0\times (\frac{1}{2})^{\gamma-1}=\frac{T_0}{2^{\gamma-1}}$ $\\$ $P_1V_1^{\gamma}=P_2V_2^{\gamma} \Rightarrow P_0V_0^{\gamma}=p_1(2V_0)^{\gamma} \Rightarrow P_1=P_0(\frac{V_0}{2V_0})^{\gamma}=\frac{P_0}{2^{\gamma}}$ $\\$ (b) When the values are opened, the temperature remains $T_0$ through out $\\$ $P_1=\frac{n_1RT}{4V_0},P_2=\frac{n_2RT_0}{4V_0}$ [Total value after the expt =$2V_0+2V_0=4V_0$] $\\$ P=$P_1+P_2=\frac{(n_1+n_2)RT_0}{4V_0}=\frac{2nRT_0}{4V_0}=\frac{nRT_0}{2V}=\frac{P_0}{2}$ $\\$

30   30. Figure (27-E7) shows an adiabatic cylindrical tube of volume $V_0$ divided in two parts by a frictionless adiabatic separator. Initially, the separator is kept in the middle, an ideal gas at pressure p, and temperature T, is injected into the left part and another ideal gas at pressure p2 and temperature T2 is injected into the right part. Cp/Cv = $\gamma$ is the same for both the gases. The separator is slid slowly and is released at a position where it can stay in equilibrium. Find (a) the volumes of the two parts, b) the heat given to the gas in the left part and (c) the final common pressure of the gases.

##### Solution :

A= $1cm^2=1\times10^{-4}m^2, M=0.03g=0.03\times10^{-3}Kg$ $\\$ P=1atm=$10^5 Pascal, L=40cm=0.4m$ $\\$ $L_1=80cm=0.8m, P=0.355 atm$ $\\$ The process is adiabatic $\\$ $P(V)^\gamma=P(V)^\gamma\Rightarrow 1\times(AL)^\gamma=0.355\times(A2L)^\gamma \Rightarrow 1\hspace{0.2cm} 1=0.355 2^\gamma \Rightarrow \frac{1}{0.355}=2^\gamma$ $\\$ $=\gamma log 2=log(\frac{1}{0.355})=1.4941$ $\\$ V=$\sqrt{\frac{\gamma P}{f}}=\sqrt{\frac{1.4941\times10^5}{m/v}}=\sqrt{\frac{1.4941\times 10^5}{(\frac{0.03\times10^{-3}}{10^{-4}\times 1 \times 0.4})}}=\sqrt{\frac{1.441\times10^5\times 4 \times 10^{-5}}{3\times 10^{-5}}}=446.33\approx 447m/s$ $\\$

31   31. An adiabatic cylindrical tube of cross-sectional area 1 cm 2 is closed at one end and fitted with a piston at the other end. The tube contains 0.03 g of an ideal gas. At 1 atm pressure and at the temperature of the surrounding, the length of the gas column is 40 cm. The piston is suddenly pulled out to double the length of the column. The pressure of the gas falls to 0.355 atm. Find the speed of sound in the gas at atmospheric temperature.

##### Solution :

A= $1cm^2=1\times10^{-4}m^2, M=0.03g=0.03\times10^{-3}Kg$ $\\$ P=1atm=$10^5 Pascal, L=40cm=0.4m$ $\\$ $L_1=80cm=0.8m, P=0.355 atm$ $\\$ The process is adiabatic $\\$ $P(V)^\gamma=P(V)^\gamma\Rightarrow 1\times(AL)^\gamma=0.355\times(A2L)^\gamma \Rightarrow 1\hspace{0.2cm} 1=0.355 2^\gamma \Rightarrow \frac{1}{0.355}=2^\gamma$ $\\$ $=\gamma log 2=log(\frac{1}{0.355})=1.4941$ $\\$ V=$\sqrt{\frac{\gamma P}{f}}=\sqrt{\frac{1.4941\times10^5}{m/v}}=\sqrt{\frac{1.4941\times 10^5}{(\frac{0.03\times10^{-3}}{10^{-4}\times 1 \times 0.4})}}=\sqrt{\frac{1.441\times10^5\times 4 \times 10^{-5}}{3\times 10^{-5}}}=446.33\approx 447m/s$ $\\$

32   32. The speed of sound in hydrogen at 0°C is 1280 in/s. The density of hydrogen at STP is 0.089 kg/m 3. Calculate the molar heat capacities CP and CV of hydrogen.

##### Solution :

$\mu=4g=4\times10^{-4} Kg, \hspace{0.3cm} V=2240 cm^3=22400\times 10^{-6}m^3$ $\\$ $C_P=5 cal/mol-ki=5\times 4.2 J/mol-k=21J/mol-K$ $\\$ $C_P=\frac{\gamma R}{\gamma-1}=\frac{\gamma\times8.3}{\gamma-1}$ $\\$ $\Rightarrow 21(\gamma-1)=\gamma(8.3) \Rightarrow 21 \gamma-21=8.3\gamma\Rightarrow\gamma=\frac{21}{12.7}$ $\\$ Since the condition is STP, P=1atm=$10^5Pa$ $\\$ V=$\sqrt{\frac{\gamma f}{f}}=\sqrt{\frac{\frac{21}{12.7}\times10^5}{\frac{4\times 10^{-3}}{2240\times10^{-6}}}}=\sqrt{\frac{21\times10^5\times 2240\times 10^{-6}}{12.7\times4\times10^{-3}}}=962.28 m/s$ $\\$

33   33. 4 g of helium occupies 22400 cm$^3$ at STP. The specific heat capacity of helium at constant pressure is 5.0 cal/mol-K. Calculate the speed of sound in helium at STP

##### Solution :

34   33. 4 g of helium occupies 22400 cm$^3$ at STP. The specific heat capacity of helium at constant pressure is 5.0 cal/mol-K. Calculate the speed of sound in helium at STP

##### Solution :

$\mu=4g=4\times10^{-4} Kg, \hspace{0.3cm} V=2240 cm^3=22400\times 10^{-6}m^3$ $\\$ $C_P=5 cal/mol-ki=5\times 4.2 J/mol-k=21J/mol-K$ $\\$ $C_P=\frac{\gamma R}{\gamma-1}=\frac{\gamma\times8.3}{\gamma-1}$ $\\$ $\Rightarrow 21(\gamma-1)=\gamma(8.3) \Rightarrow 21 \gamma-21=8.3\gamma\Rightarrow\gamma=\frac{21}{12.7}$ $\\$ Since the condition is STP, P=1atm=$10^5Pa$ $\\$ V=$\sqrt{\frac{\gamma f}{f}}=\sqrt{\frac{\frac{21}{12.7}\times10^5}{\frac{4\times 10^{-3}}{2240\times10^{-6}}}}=\sqrt{\frac{21\times10^5\times 2240\times 10^{-6}}{12.7\times4\times10^{-3}}}=962.28 m/s$ $\\$

35   34. An ideal gas having density $1.7 \times 10^{-3}$ g/$cm^3$ at a pressure 1-5 x 10° Pa is filled in a Kundt's tube. Whenthe gas is resonated at a frequency of 30 kHz, nodes are formed at a separation of 6.0 cm. Calculate the molar heat capacities Cp and Cv of the gas.

##### Solution :

Given $f_0=1.7\times10^{-4}g/cm^3=1.7kg/m^3, P=1.5\times10^5Pa, R=8.3 J/mol-k$ $\\$ $f=3.0KHZ$ $\\$ Node separation in a Kunt' tube =$\frac{\lambda}{2}=6 cm, \Rightarrow \lambda=12cm=12\times10^{-3} m$ $\\$ So, V=f$\lambda=3 \times 10^3\times 12 \times 10^{-2}=360m/s$ $\\$ We know, that the speed of sound=$\sqrt{\frac{\gamma P}{f_0}}\Rightarrow (360)^2=\frac{\gamma\times1.5\times10^5}{1.7} \Rightarrow \gamma=\frac{(360)^2\times 1.7}{1.5\times 10^5}=1.4688$ $\\$ But Cv= $\frac{R}{\gamma-1}=\frac{8.3}{1.488-1}=17.2 J/mol-k$ $\\$ Again $\frac{Cp}{Cv}=\gamma \hspace{0.35cm} So,Cp=\gamma Cv=17.72\times 1.468=26.01\approx 26 J/mol-k$ $\\$

36   35. Standing waves of frequency 5.0 kHz are produced in a tube filled with oxygen at 300 K. The separation between the consecutive nodes is 3.3 cm. Calculate the specific heat capacities Cp and Cv, of the gas.

##### Solution :

f=$5 \times 10^3 Hz,$ T=300Hz, $\frac{\lambda}{2}=3.3cm \Rightarrow \lambda=6.6\times10^{-2} m$ $\\$ V=$f\lambda=5\times 10^3\times 6.6\times 10^{-2}=(66\times5)m/s$ $\\$ V=$\frac{\lambda P}{f}[Pv=nRT\Rightarrow P=\frac{m}{mV}\times RT \Rightarrow PM=f0RT\Rightarrow \frac{P}{f_0}=\frac{RT}{M}]$ $\\$ $\sqrt{\frac{\gamma RT}{m}}(66\times5)=\sqrt{\frac{\gamma \times 8.3 \times 300}{32\times 10^{-3}}} rRightarrow (66\times5)^2=\frac{\gamma\times8.3\times300}{32\times10^{-3}} \Rightarrow \gamma=\frac{(66\times5)^2\times32\times10^{-3}}{8.3\times300}=1.3995$ $\\$ $C_v=\frac{R}{\gamma-1}=\frac{8.3}{0.3995}=20.7 J/mol-k,$ $\\$ $C_P=C_V+R=20.77+8.3=29.07 J/mol-k.$ $\\$