148466
A diatomic gas of volume $2 \mathrm{~m}^{3}$ at a pressure $2 \times$ $10^{5} \mathrm{~N} / \mathrm{m}^{2}$ is compressed adiabatically to a volume $0.5 \mathrm{~m}^{3}$. The work done in this process is. [Use $4^{1.4}=6.96$ ]
148468
One liter of a gas (with $\gamma=\frac{5}{3}$ ) at NTP is compressed adiabatically to one cubic centimeter, then the resulting pressure is :
1 $10 \mathrm{Atm}$
2 $10^{3} \mathrm{Atm}$
3 $10^{5} \mathrm{Atm}$
4 $100 \mathrm{Atm}$
Explanation:
C Given that, $\mathrm{P}_{1}=1 \mathrm{~atm},$ $\mathrm{~V}_{1}=1 \mathrm{~L}=1000 \mathrm{~cm}^{3}$ We know that, For adiabatic process $\mathrm{P}_{1} \mathrm{~V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{~V}_{2}^{\gamma}$ $\mathrm{P}_{2}=?, \mathrm{~V}_{2}=1 \mathrm{~cm}^{3}, \gamma=\frac{5}{3}$ $(1) \times(1000)^{5 / 3}=\mathrm{P}_{2} \times(1)^{5 / 3}$ $(1000)^{5 / 3}=\mathrm{P}_{2}$ $\mathrm{P}_{2}=10^{5} \mathrm{~atm}$
MP PMT-2013
Thermodynamics
148469
For a gas of non rigid diatomic molecules, the value of $\gamma=\frac{C_{p}}{C_{v}}$ is :
1 $\frac{9}{7}$
2 $\frac{7}{5}$
3 $\frac{5}{3}$
4 $\frac{11}{9}$
Explanation:
B The value of $\gamma$ for diatomic $\gamma=1+\frac{2}{\mathrm{f}}$ Degree of freedom for diatomic $(\mathrm{f})=5$ Hence, $\quad \gamma=1+\frac{2}{\mathrm{f}}$ $\gamma=1+\frac{2}{5}$ $\gamma=7 / 5$
MP PMT-2013
Thermodynamics
148470
P-V plots for gases during adiabatic process as shown in figure plot 1 and 2 should correspond respectively to
1 $\mathrm{He}$ and $\mathrm{O}_{2}$
2 $\mathrm{O}_{2}$ and $\mathrm{He}$
3 $\mathrm{He}$ and $\mathrm{Ar}$
4 $\mathrm{O}_{2}$ and $\mathrm{N}_{2}$
Explanation:
B The slope of adiabatic process of $\mathrm{P}-\mathrm{V}$ $\text { diagram }\left(\frac{\mathrm{dP}}{\mathrm{dV}}\right)=\gamma \frac{\mathrm{P}}{\mathrm{V}}$ $\text { From graph (slope })_{2}>(\text { slope })_{1}$ $\Rightarrow \gamma_{2}>\gamma_{1}$ $\gamma \text { for monoatomic gas }(\mathrm{Ar}, \mathrm{He}) \text { is greater than } \gamma \text { for }$ $\text { diatomic gas }\left(\mathrm{O}_{2}, \mathrm{~N}_{2}\right) \text {. }$
148466
A diatomic gas of volume $2 \mathrm{~m}^{3}$ at a pressure $2 \times$ $10^{5} \mathrm{~N} / \mathrm{m}^{2}$ is compressed adiabatically to a volume $0.5 \mathrm{~m}^{3}$. The work done in this process is. [Use $4^{1.4}=6.96$ ]
148468
One liter of a gas (with $\gamma=\frac{5}{3}$ ) at NTP is compressed adiabatically to one cubic centimeter, then the resulting pressure is :
1 $10 \mathrm{Atm}$
2 $10^{3} \mathrm{Atm}$
3 $10^{5} \mathrm{Atm}$
4 $100 \mathrm{Atm}$
Explanation:
C Given that, $\mathrm{P}_{1}=1 \mathrm{~atm},$ $\mathrm{~V}_{1}=1 \mathrm{~L}=1000 \mathrm{~cm}^{3}$ We know that, For adiabatic process $\mathrm{P}_{1} \mathrm{~V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{~V}_{2}^{\gamma}$ $\mathrm{P}_{2}=?, \mathrm{~V}_{2}=1 \mathrm{~cm}^{3}, \gamma=\frac{5}{3}$ $(1) \times(1000)^{5 / 3}=\mathrm{P}_{2} \times(1)^{5 / 3}$ $(1000)^{5 / 3}=\mathrm{P}_{2}$ $\mathrm{P}_{2}=10^{5} \mathrm{~atm}$
MP PMT-2013
Thermodynamics
148469
For a gas of non rigid diatomic molecules, the value of $\gamma=\frac{C_{p}}{C_{v}}$ is :
1 $\frac{9}{7}$
2 $\frac{7}{5}$
3 $\frac{5}{3}$
4 $\frac{11}{9}$
Explanation:
B The value of $\gamma$ for diatomic $\gamma=1+\frac{2}{\mathrm{f}}$ Degree of freedom for diatomic $(\mathrm{f})=5$ Hence, $\quad \gamma=1+\frac{2}{\mathrm{f}}$ $\gamma=1+\frac{2}{5}$ $\gamma=7 / 5$
MP PMT-2013
Thermodynamics
148470
P-V plots for gases during adiabatic process as shown in figure plot 1 and 2 should correspond respectively to
1 $\mathrm{He}$ and $\mathrm{O}_{2}$
2 $\mathrm{O}_{2}$ and $\mathrm{He}$
3 $\mathrm{He}$ and $\mathrm{Ar}$
4 $\mathrm{O}_{2}$ and $\mathrm{N}_{2}$
Explanation:
B The slope of adiabatic process of $\mathrm{P}-\mathrm{V}$ $\text { diagram }\left(\frac{\mathrm{dP}}{\mathrm{dV}}\right)=\gamma \frac{\mathrm{P}}{\mathrm{V}}$ $\text { From graph (slope })_{2}>(\text { slope })_{1}$ $\Rightarrow \gamma_{2}>\gamma_{1}$ $\gamma \text { for monoatomic gas }(\mathrm{Ar}, \mathrm{He}) \text { is greater than } \gamma \text { for }$ $\text { diatomic gas }\left(\mathrm{O}_{2}, \mathrm{~N}_{2}\right) \text {. }$
148466
A diatomic gas of volume $2 \mathrm{~m}^{3}$ at a pressure $2 \times$ $10^{5} \mathrm{~N} / \mathrm{m}^{2}$ is compressed adiabatically to a volume $0.5 \mathrm{~m}^{3}$. The work done in this process is. [Use $4^{1.4}=6.96$ ]
148468
One liter of a gas (with $\gamma=\frac{5}{3}$ ) at NTP is compressed adiabatically to one cubic centimeter, then the resulting pressure is :
1 $10 \mathrm{Atm}$
2 $10^{3} \mathrm{Atm}$
3 $10^{5} \mathrm{Atm}$
4 $100 \mathrm{Atm}$
Explanation:
C Given that, $\mathrm{P}_{1}=1 \mathrm{~atm},$ $\mathrm{~V}_{1}=1 \mathrm{~L}=1000 \mathrm{~cm}^{3}$ We know that, For adiabatic process $\mathrm{P}_{1} \mathrm{~V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{~V}_{2}^{\gamma}$ $\mathrm{P}_{2}=?, \mathrm{~V}_{2}=1 \mathrm{~cm}^{3}, \gamma=\frac{5}{3}$ $(1) \times(1000)^{5 / 3}=\mathrm{P}_{2} \times(1)^{5 / 3}$ $(1000)^{5 / 3}=\mathrm{P}_{2}$ $\mathrm{P}_{2}=10^{5} \mathrm{~atm}$
MP PMT-2013
Thermodynamics
148469
For a gas of non rigid diatomic molecules, the value of $\gamma=\frac{C_{p}}{C_{v}}$ is :
1 $\frac{9}{7}$
2 $\frac{7}{5}$
3 $\frac{5}{3}$
4 $\frac{11}{9}$
Explanation:
B The value of $\gamma$ for diatomic $\gamma=1+\frac{2}{\mathrm{f}}$ Degree of freedom for diatomic $(\mathrm{f})=5$ Hence, $\quad \gamma=1+\frac{2}{\mathrm{f}}$ $\gamma=1+\frac{2}{5}$ $\gamma=7 / 5$
MP PMT-2013
Thermodynamics
148470
P-V plots for gases during adiabatic process as shown in figure plot 1 and 2 should correspond respectively to
1 $\mathrm{He}$ and $\mathrm{O}_{2}$
2 $\mathrm{O}_{2}$ and $\mathrm{He}$
3 $\mathrm{He}$ and $\mathrm{Ar}$
4 $\mathrm{O}_{2}$ and $\mathrm{N}_{2}$
Explanation:
B The slope of adiabatic process of $\mathrm{P}-\mathrm{V}$ $\text { diagram }\left(\frac{\mathrm{dP}}{\mathrm{dV}}\right)=\gamma \frac{\mathrm{P}}{\mathrm{V}}$ $\text { From graph (slope })_{2}>(\text { slope })_{1}$ $\Rightarrow \gamma_{2}>\gamma_{1}$ $\gamma \text { for monoatomic gas }(\mathrm{Ar}, \mathrm{He}) \text { is greater than } \gamma \text { for }$ $\text { diatomic gas }\left(\mathrm{O}_{2}, \mathrm{~N}_{2}\right) \text {. }$
148466
A diatomic gas of volume $2 \mathrm{~m}^{3}$ at a pressure $2 \times$ $10^{5} \mathrm{~N} / \mathrm{m}^{2}$ is compressed adiabatically to a volume $0.5 \mathrm{~m}^{3}$. The work done in this process is. [Use $4^{1.4}=6.96$ ]
148468
One liter of a gas (with $\gamma=\frac{5}{3}$ ) at NTP is compressed adiabatically to one cubic centimeter, then the resulting pressure is :
1 $10 \mathrm{Atm}$
2 $10^{3} \mathrm{Atm}$
3 $10^{5} \mathrm{Atm}$
4 $100 \mathrm{Atm}$
Explanation:
C Given that, $\mathrm{P}_{1}=1 \mathrm{~atm},$ $\mathrm{~V}_{1}=1 \mathrm{~L}=1000 \mathrm{~cm}^{3}$ We know that, For adiabatic process $\mathrm{P}_{1} \mathrm{~V}_{1}^{\gamma}=\mathrm{P}_{2} \mathrm{~V}_{2}^{\gamma}$ $\mathrm{P}_{2}=?, \mathrm{~V}_{2}=1 \mathrm{~cm}^{3}, \gamma=\frac{5}{3}$ $(1) \times(1000)^{5 / 3}=\mathrm{P}_{2} \times(1)^{5 / 3}$ $(1000)^{5 / 3}=\mathrm{P}_{2}$ $\mathrm{P}_{2}=10^{5} \mathrm{~atm}$
MP PMT-2013
Thermodynamics
148469
For a gas of non rigid diatomic molecules, the value of $\gamma=\frac{C_{p}}{C_{v}}$ is :
1 $\frac{9}{7}$
2 $\frac{7}{5}$
3 $\frac{5}{3}$
4 $\frac{11}{9}$
Explanation:
B The value of $\gamma$ for diatomic $\gamma=1+\frac{2}{\mathrm{f}}$ Degree of freedom for diatomic $(\mathrm{f})=5$ Hence, $\quad \gamma=1+\frac{2}{\mathrm{f}}$ $\gamma=1+\frac{2}{5}$ $\gamma=7 / 5$
MP PMT-2013
Thermodynamics
148470
P-V plots for gases during adiabatic process as shown in figure plot 1 and 2 should correspond respectively to
1 $\mathrm{He}$ and $\mathrm{O}_{2}$
2 $\mathrm{O}_{2}$ and $\mathrm{He}$
3 $\mathrm{He}$ and $\mathrm{Ar}$
4 $\mathrm{O}_{2}$ and $\mathrm{N}_{2}$
Explanation:
B The slope of adiabatic process of $\mathrm{P}-\mathrm{V}$ $\text { diagram }\left(\frac{\mathrm{dP}}{\mathrm{dV}}\right)=\gamma \frac{\mathrm{P}}{\mathrm{V}}$ $\text { From graph (slope })_{2}>(\text { slope })_{1}$ $\Rightarrow \gamma_{2}>\gamma_{1}$ $\gamma \text { for monoatomic gas }(\mathrm{Ar}, \mathrm{He}) \text { is greater than } \gamma \text { for }$ $\text { diatomic gas }\left(\mathrm{O}_{2}, \mathrm{~N}_{2}\right) \text {. }$
