148382
In an isobaric process of an ideal gas. The ratio of heat supplied and work done by the system $\left[\right.$ i.e. $\left.\left(\frac{\mathbf{Q}}{\mathbf{W}}\right)\right]$ is
1 $\frac{\gamma-1}{\gamma}$
2 $\gamma$
3 $\frac{\gamma}{\gamma-1}$
4 1
Explanation:
C Given, The process is Isobaric Now, for Isobaric process $\mathrm{Q}=\mathrm{mC}_{\mathrm{P}} \mathrm{dT}$ $\mathrm{W}=\mathrm{PdV}$ We know that, $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma$ From equation (iii) and (iv), we get $\mathrm{C}_{\mathrm{P}}=\frac{\gamma \mathrm{R}}{\gamma-1}$ $\therefore \frac{\mathrm{Q}}{\mathrm{W}}=\frac{\mathrm{mC}_{\mathrm{P}} \mathrm{dT}}{\mathrm{PdV}}$ $=\frac{\mathrm{m} \times\left(\frac{\gamma \mathrm{R}}{\gamma-1}\right) \mathrm{dT}}{\mathrm{PdV}}$ $=\left(\frac{\gamma}{\gamma-1}\right) \frac{\mathrm{mRdT}}{\mathrm{PdV}}$ For Ideal gas $\mathrm{PV}=\mathrm{mRT}$ $\therefore \frac{\mathrm{Q}}{\mathrm{W}}=\frac{\gamma}{\gamma-1}\left(\frac{\mathrm{PdV}}{\mathrm{PdV}}\right)$ $\frac{\mathrm{Q}}{\mathrm{W}}=\frac{\gamma}{\gamma-1}$
AMU-2001
Thermodynamics
148384
The temperature of a gas contained in a closed vessel increases by $2^{\circ} \mathrm{C}$ when the pressure is increased by $2 \%$. The initial temperature of the gas is:
1 $200 \mathrm{~K}$
2 $100 \mathrm{~K}$
3 $200^{\circ} \mathrm{C}$
4 $100^{\circ} \mathrm{C}$
Explanation:
B Given, Temperature of a gas increased $(\Delta \mathrm{T})=2^{\circ} \mathrm{C}$ Pressure increased $=2 \%=\frac{2}{100}=0.02$ We know that, Ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ According to the question Let, Initial temperature $\left(\mathrm{T}_{1}\right)=\mathrm{T}$ Final temperature $\left(\mathrm{T}_{2}\right)=\mathrm{T}+2$ Initial pressure $\left(\mathrm{P}_{1}\right)=\mathrm{P}$ Final pressure $\left(\mathrm{P}_{2}\right)=1.02 \mathrm{P}$ Now from Ideal gas equation $\frac{\mathrm{P}}{\mathrm{T}}=$ constant $\frac{\mathrm{P}}{\mathrm{T}}=\frac{1.02 \mathrm{P}}{\mathrm{T}+2}$ $\frac{\mathrm{T}+2}{\mathrm{~T}}=1.02$ $1+\frac{2}{\mathrm{~T}}=1.02$ $\frac{2}{\mathrm{~T}}=0.02$ $\mathrm{T}=\frac{200}{2}$ $\mathrm{T}=100 \mathrm{~K}$ $\therefore$ Initial temperature of gas $=100 \mathrm{~K}$.
AP EAMCET(Medical)-2006
Thermodynamics
148386
When heat energy of $1500 \mathrm{~J}$ is supplied to a gas at constant pressure, $2.1 \times 10^{5} \mathrm{Nm}^{-2}$, there was an increase in its volume equal to $2.5 \times 10^{-3} \mathrm{~m}^{3}$. The increase in its internal energy in joule is
1 $450 \mathrm{~J}$
2 $525 \mathrm{~J}$
3 $975 \mathrm{~J}$
4 $2025 \mathrm{~J}$
Explanation:
C Given, Heat energy supplied to a gas $(\mathrm{Q})=1500 \mathrm{~J}$ Pressure (remains constant $)=2.1 \times 10^{5} \mathrm{Nm}^{-2}$ Increase in its volume $\quad=2.5 \times 10^{-3} \mathrm{~m}^{3}$ We know that, $\mathrm{I}^{\text {st }}$ law of thermodynamics, $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$ $\therefore \quad \Delta \mathrm{W}=\mathrm{PdV}$ $=2.1 \times 10^{5}\left(2.5 \times 10^{-3}\right)$ $=21 \times 25$ $=525 \mathrm{~J}$ $\therefore \quad \Delta \mathrm{U}=\Delta \mathrm{Q}-\Delta \mathrm{W}$ $=1500-525$ $\Delta \mathrm{U}=975 \mathrm{~J}$
EAMCET-1999
Thermodynamics
148387
When an ideal diatomic gas is heated at constant pressure, fraction of heat energy supplied that increases the internal energy of the gas is
1 $\frac{5}{7}$
2 $\frac{7}{5}$
3 $\frac{3}{5}$
4 $\frac{5}{3}$
5 $\frac{2}{3}$
Explanation:
A Given, For Ideal diatomic gas $(\gamma)=1+2 / \mathrm{f}=1+2 / 5=7 / 5$ $\therefore \quad \mathrm{C}_{\mathrm{V}}=\frac{\mathrm{R}}{\gamma-1}=\frac{5}{2} \mathrm{R}$ $\mathrm{C}_{\mathrm{P}}=\frac{\gamma \mathrm{R}}{\gamma-1}=\frac{7}{2} \mathrm{R}$ Fraction of heat energy which increase internal energy $\left(\frac{\mathrm{dU}}{\mathrm{dQ}}\right)_{\mathrm{P}}=\frac{\mathrm{nC}_{\mathrm{V}} \mathrm{dT}}{\mathrm{nC}_{\mathrm{P}} \mathrm{dT}}=\frac{\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{P}}}=\frac{5 / 2 \mathrm{R}}{7 / 2 \mathrm{R}}=\frac{5}{7}$
148382
In an isobaric process of an ideal gas. The ratio of heat supplied and work done by the system $\left[\right.$ i.e. $\left.\left(\frac{\mathbf{Q}}{\mathbf{W}}\right)\right]$ is
1 $\frac{\gamma-1}{\gamma}$
2 $\gamma$
3 $\frac{\gamma}{\gamma-1}$
4 1
Explanation:
C Given, The process is Isobaric Now, for Isobaric process $\mathrm{Q}=\mathrm{mC}_{\mathrm{P}} \mathrm{dT}$ $\mathrm{W}=\mathrm{PdV}$ We know that, $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma$ From equation (iii) and (iv), we get $\mathrm{C}_{\mathrm{P}}=\frac{\gamma \mathrm{R}}{\gamma-1}$ $\therefore \frac{\mathrm{Q}}{\mathrm{W}}=\frac{\mathrm{mC}_{\mathrm{P}} \mathrm{dT}}{\mathrm{PdV}}$ $=\frac{\mathrm{m} \times\left(\frac{\gamma \mathrm{R}}{\gamma-1}\right) \mathrm{dT}}{\mathrm{PdV}}$ $=\left(\frac{\gamma}{\gamma-1}\right) \frac{\mathrm{mRdT}}{\mathrm{PdV}}$ For Ideal gas $\mathrm{PV}=\mathrm{mRT}$ $\therefore \frac{\mathrm{Q}}{\mathrm{W}}=\frac{\gamma}{\gamma-1}\left(\frac{\mathrm{PdV}}{\mathrm{PdV}}\right)$ $\frac{\mathrm{Q}}{\mathrm{W}}=\frac{\gamma}{\gamma-1}$
AMU-2001
Thermodynamics
148384
The temperature of a gas contained in a closed vessel increases by $2^{\circ} \mathrm{C}$ when the pressure is increased by $2 \%$. The initial temperature of the gas is:
1 $200 \mathrm{~K}$
2 $100 \mathrm{~K}$
3 $200^{\circ} \mathrm{C}$
4 $100^{\circ} \mathrm{C}$
Explanation:
B Given, Temperature of a gas increased $(\Delta \mathrm{T})=2^{\circ} \mathrm{C}$ Pressure increased $=2 \%=\frac{2}{100}=0.02$ We know that, Ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ According to the question Let, Initial temperature $\left(\mathrm{T}_{1}\right)=\mathrm{T}$ Final temperature $\left(\mathrm{T}_{2}\right)=\mathrm{T}+2$ Initial pressure $\left(\mathrm{P}_{1}\right)=\mathrm{P}$ Final pressure $\left(\mathrm{P}_{2}\right)=1.02 \mathrm{P}$ Now from Ideal gas equation $\frac{\mathrm{P}}{\mathrm{T}}=$ constant $\frac{\mathrm{P}}{\mathrm{T}}=\frac{1.02 \mathrm{P}}{\mathrm{T}+2}$ $\frac{\mathrm{T}+2}{\mathrm{~T}}=1.02$ $1+\frac{2}{\mathrm{~T}}=1.02$ $\frac{2}{\mathrm{~T}}=0.02$ $\mathrm{T}=\frac{200}{2}$ $\mathrm{T}=100 \mathrm{~K}$ $\therefore$ Initial temperature of gas $=100 \mathrm{~K}$.
AP EAMCET(Medical)-2006
Thermodynamics
148386
When heat energy of $1500 \mathrm{~J}$ is supplied to a gas at constant pressure, $2.1 \times 10^{5} \mathrm{Nm}^{-2}$, there was an increase in its volume equal to $2.5 \times 10^{-3} \mathrm{~m}^{3}$. The increase in its internal energy in joule is
1 $450 \mathrm{~J}$
2 $525 \mathrm{~J}$
3 $975 \mathrm{~J}$
4 $2025 \mathrm{~J}$
Explanation:
C Given, Heat energy supplied to a gas $(\mathrm{Q})=1500 \mathrm{~J}$ Pressure (remains constant $)=2.1 \times 10^{5} \mathrm{Nm}^{-2}$ Increase in its volume $\quad=2.5 \times 10^{-3} \mathrm{~m}^{3}$ We know that, $\mathrm{I}^{\text {st }}$ law of thermodynamics, $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$ $\therefore \quad \Delta \mathrm{W}=\mathrm{PdV}$ $=2.1 \times 10^{5}\left(2.5 \times 10^{-3}\right)$ $=21 \times 25$ $=525 \mathrm{~J}$ $\therefore \quad \Delta \mathrm{U}=\Delta \mathrm{Q}-\Delta \mathrm{W}$ $=1500-525$ $\Delta \mathrm{U}=975 \mathrm{~J}$
EAMCET-1999
Thermodynamics
148387
When an ideal diatomic gas is heated at constant pressure, fraction of heat energy supplied that increases the internal energy of the gas is
1 $\frac{5}{7}$
2 $\frac{7}{5}$
3 $\frac{3}{5}$
4 $\frac{5}{3}$
5 $\frac{2}{3}$
Explanation:
A Given, For Ideal diatomic gas $(\gamma)=1+2 / \mathrm{f}=1+2 / 5=7 / 5$ $\therefore \quad \mathrm{C}_{\mathrm{V}}=\frac{\mathrm{R}}{\gamma-1}=\frac{5}{2} \mathrm{R}$ $\mathrm{C}_{\mathrm{P}}=\frac{\gamma \mathrm{R}}{\gamma-1}=\frac{7}{2} \mathrm{R}$ Fraction of heat energy which increase internal energy $\left(\frac{\mathrm{dU}}{\mathrm{dQ}}\right)_{\mathrm{P}}=\frac{\mathrm{nC}_{\mathrm{V}} \mathrm{dT}}{\mathrm{nC}_{\mathrm{P}} \mathrm{dT}}=\frac{\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{P}}}=\frac{5 / 2 \mathrm{R}}{7 / 2 \mathrm{R}}=\frac{5}{7}$
148382
In an isobaric process of an ideal gas. The ratio of heat supplied and work done by the system $\left[\right.$ i.e. $\left.\left(\frac{\mathbf{Q}}{\mathbf{W}}\right)\right]$ is
1 $\frac{\gamma-1}{\gamma}$
2 $\gamma$
3 $\frac{\gamma}{\gamma-1}$
4 1
Explanation:
C Given, The process is Isobaric Now, for Isobaric process $\mathrm{Q}=\mathrm{mC}_{\mathrm{P}} \mathrm{dT}$ $\mathrm{W}=\mathrm{PdV}$ We know that, $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma$ From equation (iii) and (iv), we get $\mathrm{C}_{\mathrm{P}}=\frac{\gamma \mathrm{R}}{\gamma-1}$ $\therefore \frac{\mathrm{Q}}{\mathrm{W}}=\frac{\mathrm{mC}_{\mathrm{P}} \mathrm{dT}}{\mathrm{PdV}}$ $=\frac{\mathrm{m} \times\left(\frac{\gamma \mathrm{R}}{\gamma-1}\right) \mathrm{dT}}{\mathrm{PdV}}$ $=\left(\frac{\gamma}{\gamma-1}\right) \frac{\mathrm{mRdT}}{\mathrm{PdV}}$ For Ideal gas $\mathrm{PV}=\mathrm{mRT}$ $\therefore \frac{\mathrm{Q}}{\mathrm{W}}=\frac{\gamma}{\gamma-1}\left(\frac{\mathrm{PdV}}{\mathrm{PdV}}\right)$ $\frac{\mathrm{Q}}{\mathrm{W}}=\frac{\gamma}{\gamma-1}$
AMU-2001
Thermodynamics
148384
The temperature of a gas contained in a closed vessel increases by $2^{\circ} \mathrm{C}$ when the pressure is increased by $2 \%$. The initial temperature of the gas is:
1 $200 \mathrm{~K}$
2 $100 \mathrm{~K}$
3 $200^{\circ} \mathrm{C}$
4 $100^{\circ} \mathrm{C}$
Explanation:
B Given, Temperature of a gas increased $(\Delta \mathrm{T})=2^{\circ} \mathrm{C}$ Pressure increased $=2 \%=\frac{2}{100}=0.02$ We know that, Ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ According to the question Let, Initial temperature $\left(\mathrm{T}_{1}\right)=\mathrm{T}$ Final temperature $\left(\mathrm{T}_{2}\right)=\mathrm{T}+2$ Initial pressure $\left(\mathrm{P}_{1}\right)=\mathrm{P}$ Final pressure $\left(\mathrm{P}_{2}\right)=1.02 \mathrm{P}$ Now from Ideal gas equation $\frac{\mathrm{P}}{\mathrm{T}}=$ constant $\frac{\mathrm{P}}{\mathrm{T}}=\frac{1.02 \mathrm{P}}{\mathrm{T}+2}$ $\frac{\mathrm{T}+2}{\mathrm{~T}}=1.02$ $1+\frac{2}{\mathrm{~T}}=1.02$ $\frac{2}{\mathrm{~T}}=0.02$ $\mathrm{T}=\frac{200}{2}$ $\mathrm{T}=100 \mathrm{~K}$ $\therefore$ Initial temperature of gas $=100 \mathrm{~K}$.
AP EAMCET(Medical)-2006
Thermodynamics
148386
When heat energy of $1500 \mathrm{~J}$ is supplied to a gas at constant pressure, $2.1 \times 10^{5} \mathrm{Nm}^{-2}$, there was an increase in its volume equal to $2.5 \times 10^{-3} \mathrm{~m}^{3}$. The increase in its internal energy in joule is
1 $450 \mathrm{~J}$
2 $525 \mathrm{~J}$
3 $975 \mathrm{~J}$
4 $2025 \mathrm{~J}$
Explanation:
C Given, Heat energy supplied to a gas $(\mathrm{Q})=1500 \mathrm{~J}$ Pressure (remains constant $)=2.1 \times 10^{5} \mathrm{Nm}^{-2}$ Increase in its volume $\quad=2.5 \times 10^{-3} \mathrm{~m}^{3}$ We know that, $\mathrm{I}^{\text {st }}$ law of thermodynamics, $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$ $\therefore \quad \Delta \mathrm{W}=\mathrm{PdV}$ $=2.1 \times 10^{5}\left(2.5 \times 10^{-3}\right)$ $=21 \times 25$ $=525 \mathrm{~J}$ $\therefore \quad \Delta \mathrm{U}=\Delta \mathrm{Q}-\Delta \mathrm{W}$ $=1500-525$ $\Delta \mathrm{U}=975 \mathrm{~J}$
EAMCET-1999
Thermodynamics
148387
When an ideal diatomic gas is heated at constant pressure, fraction of heat energy supplied that increases the internal energy of the gas is
1 $\frac{5}{7}$
2 $\frac{7}{5}$
3 $\frac{3}{5}$
4 $\frac{5}{3}$
5 $\frac{2}{3}$
Explanation:
A Given, For Ideal diatomic gas $(\gamma)=1+2 / \mathrm{f}=1+2 / 5=7 / 5$ $\therefore \quad \mathrm{C}_{\mathrm{V}}=\frac{\mathrm{R}}{\gamma-1}=\frac{5}{2} \mathrm{R}$ $\mathrm{C}_{\mathrm{P}}=\frac{\gamma \mathrm{R}}{\gamma-1}=\frac{7}{2} \mathrm{R}$ Fraction of heat energy which increase internal energy $\left(\frac{\mathrm{dU}}{\mathrm{dQ}}\right)_{\mathrm{P}}=\frac{\mathrm{nC}_{\mathrm{V}} \mathrm{dT}}{\mathrm{nC}_{\mathrm{P}} \mathrm{dT}}=\frac{\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{P}}}=\frac{5 / 2 \mathrm{R}}{7 / 2 \mathrm{R}}=\frac{5}{7}$
148382
In an isobaric process of an ideal gas. The ratio of heat supplied and work done by the system $\left[\right.$ i.e. $\left.\left(\frac{\mathbf{Q}}{\mathbf{W}}\right)\right]$ is
1 $\frac{\gamma-1}{\gamma}$
2 $\gamma$
3 $\frac{\gamma}{\gamma-1}$
4 1
Explanation:
C Given, The process is Isobaric Now, for Isobaric process $\mathrm{Q}=\mathrm{mC}_{\mathrm{P}} \mathrm{dT}$ $\mathrm{W}=\mathrm{PdV}$ We know that, $\mathrm{C}_{\mathrm{P}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$ $\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma$ From equation (iii) and (iv), we get $\mathrm{C}_{\mathrm{P}}=\frac{\gamma \mathrm{R}}{\gamma-1}$ $\therefore \frac{\mathrm{Q}}{\mathrm{W}}=\frac{\mathrm{mC}_{\mathrm{P}} \mathrm{dT}}{\mathrm{PdV}}$ $=\frac{\mathrm{m} \times\left(\frac{\gamma \mathrm{R}}{\gamma-1}\right) \mathrm{dT}}{\mathrm{PdV}}$ $=\left(\frac{\gamma}{\gamma-1}\right) \frac{\mathrm{mRdT}}{\mathrm{PdV}}$ For Ideal gas $\mathrm{PV}=\mathrm{mRT}$ $\therefore \frac{\mathrm{Q}}{\mathrm{W}}=\frac{\gamma}{\gamma-1}\left(\frac{\mathrm{PdV}}{\mathrm{PdV}}\right)$ $\frac{\mathrm{Q}}{\mathrm{W}}=\frac{\gamma}{\gamma-1}$
AMU-2001
Thermodynamics
148384
The temperature of a gas contained in a closed vessel increases by $2^{\circ} \mathrm{C}$ when the pressure is increased by $2 \%$. The initial temperature of the gas is:
1 $200 \mathrm{~K}$
2 $100 \mathrm{~K}$
3 $200^{\circ} \mathrm{C}$
4 $100^{\circ} \mathrm{C}$
Explanation:
B Given, Temperature of a gas increased $(\Delta \mathrm{T})=2^{\circ} \mathrm{C}$ Pressure increased $=2 \%=\frac{2}{100}=0.02$ We know that, Ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ According to the question Let, Initial temperature $\left(\mathrm{T}_{1}\right)=\mathrm{T}$ Final temperature $\left(\mathrm{T}_{2}\right)=\mathrm{T}+2$ Initial pressure $\left(\mathrm{P}_{1}\right)=\mathrm{P}$ Final pressure $\left(\mathrm{P}_{2}\right)=1.02 \mathrm{P}$ Now from Ideal gas equation $\frac{\mathrm{P}}{\mathrm{T}}=$ constant $\frac{\mathrm{P}}{\mathrm{T}}=\frac{1.02 \mathrm{P}}{\mathrm{T}+2}$ $\frac{\mathrm{T}+2}{\mathrm{~T}}=1.02$ $1+\frac{2}{\mathrm{~T}}=1.02$ $\frac{2}{\mathrm{~T}}=0.02$ $\mathrm{T}=\frac{200}{2}$ $\mathrm{T}=100 \mathrm{~K}$ $\therefore$ Initial temperature of gas $=100 \mathrm{~K}$.
AP EAMCET(Medical)-2006
Thermodynamics
148386
When heat energy of $1500 \mathrm{~J}$ is supplied to a gas at constant pressure, $2.1 \times 10^{5} \mathrm{Nm}^{-2}$, there was an increase in its volume equal to $2.5 \times 10^{-3} \mathrm{~m}^{3}$. The increase in its internal energy in joule is
1 $450 \mathrm{~J}$
2 $525 \mathrm{~J}$
3 $975 \mathrm{~J}$
4 $2025 \mathrm{~J}$
Explanation:
C Given, Heat energy supplied to a gas $(\mathrm{Q})=1500 \mathrm{~J}$ Pressure (remains constant $)=2.1 \times 10^{5} \mathrm{Nm}^{-2}$ Increase in its volume $\quad=2.5 \times 10^{-3} \mathrm{~m}^{3}$ We know that, $\mathrm{I}^{\text {st }}$ law of thermodynamics, $\Delta \mathrm{Q}=\Delta \mathrm{U}+\Delta \mathrm{W}$ $\therefore \quad \Delta \mathrm{W}=\mathrm{PdV}$ $=2.1 \times 10^{5}\left(2.5 \times 10^{-3}\right)$ $=21 \times 25$ $=525 \mathrm{~J}$ $\therefore \quad \Delta \mathrm{U}=\Delta \mathrm{Q}-\Delta \mathrm{W}$ $=1500-525$ $\Delta \mathrm{U}=975 \mathrm{~J}$
EAMCET-1999
Thermodynamics
148387
When an ideal diatomic gas is heated at constant pressure, fraction of heat energy supplied that increases the internal energy of the gas is
1 $\frac{5}{7}$
2 $\frac{7}{5}$
3 $\frac{3}{5}$
4 $\frac{5}{3}$
5 $\frac{2}{3}$
Explanation:
A Given, For Ideal diatomic gas $(\gamma)=1+2 / \mathrm{f}=1+2 / 5=7 / 5$ $\therefore \quad \mathrm{C}_{\mathrm{V}}=\frac{\mathrm{R}}{\gamma-1}=\frac{5}{2} \mathrm{R}$ $\mathrm{C}_{\mathrm{P}}=\frac{\gamma \mathrm{R}}{\gamma-1}=\frac{7}{2} \mathrm{R}$ Fraction of heat energy which increase internal energy $\left(\frac{\mathrm{dU}}{\mathrm{dQ}}\right)_{\mathrm{P}}=\frac{\mathrm{nC}_{\mathrm{V}} \mathrm{dT}}{\mathrm{nC}_{\mathrm{P}} \mathrm{dT}}=\frac{\mathrm{C}_{\mathrm{V}}}{\mathrm{C}_{\mathrm{P}}}=\frac{5 / 2 \mathrm{R}}{7 / 2 \mathrm{R}}=\frac{5}{7}$
