MHT CET · Physics · Thermodynamics
The change in the internal energy of the mass of gas, when the volume changes from ' V ' to ' 2 V ' at constant pressure ' P ' is \((\gamma\) is the ratio of specific heat of gas at constant pressure to specific heat at constant volume)
- A \(\frac{\mathrm{PV}}{\gamma-1}\)
- B \(\frac{\mathrm{PV}}{\gamma+1}\)
- C \(\frac{\gamma-1}{\mathrm{PV}}\)
- D \(\frac{\gamma+1}{\mathrm{PV}}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{PV}}{\gamma-1}\)
Step-by-step Solution
Detailed explanation
Change in internal energy is given by,
\(\Delta \mathrm{U}=\mathrm{n} \mathrm{C}_{\mathrm{v}} \Delta \mathrm{~T}...(i)\)
Given, \(\frac{C_P}{C_v}=\gamma\) and \(C_P-C_v=R\)
\(1+\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{v}}}=\gamma\)
\(\begin{aligned}
C_V & =\frac{R}{(\gamma-1)} \\
\therefore \quad \Delta U & =n\left(\frac{R}{\gamma-1}\right) \Delta T
\end{aligned}\)
...[From(i)]
Using \(\mathrm{P} \Delta \mathrm{v}=\mathrm{nR} \Delta \mathrm{T}\) under constant P ,
\(\Delta U=\frac{P \Delta v}{\gamma-1}=\frac{P(2 v-v)}{\gamma-1}=\frac{P v}{(\gamma-1)}\)
\(\Delta \mathrm{U}=\mathrm{n} \mathrm{C}_{\mathrm{v}} \Delta \mathrm{~T}...(i)\)
Given, \(\frac{C_P}{C_v}=\gamma\) and \(C_P-C_v=R\)
\(1+\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{v}}}=\gamma\)
\(\begin{aligned}
C_V & =\frac{R}{(\gamma-1)} \\
\therefore \quad \Delta U & =n\left(\frac{R}{\gamma-1}\right) \Delta T
\end{aligned}\)
...[From(i)]
Using \(\mathrm{P} \Delta \mathrm{v}=\mathrm{nR} \Delta \mathrm{T}\) under constant P ,
\(\Delta U=\frac{P \Delta v}{\gamma-1}=\frac{P(2 v-v)}{\gamma-1}=\frac{P v}{(\gamma-1)}\)
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