An ideal gas, Cv = 5/2 R, is expanded adiabatically against a constant pressure of 1 atm untill it doubles in volume.

Question

An ideal gas, \(\bar{C}_v = \frac {5}{2}R,\) is expanded adiabatically against a constant pressure of 1 atm untill it doubles in volume. If the initial temperature and pressure is 298 K and 5 atm, respectively then the final temperature is ______ K (nearest integer).

[\(\bar {C}_v\) is the molar heat capacity at constant volume]

Answer 1

Correct answer is : 274

\(\Delta \mathrm{U}=\mathrm{q}+\mathrm{w}(\mathrm{q}=0) \)

\(\mathrm{nC}_{\mathrm{V}} \Delta \mathrm{T}=-\mathrm{P}_{\mathrm{cxt}}\left(\mathrm{V}_2-\mathrm{V}_1\right) \)

\(\mathrm{V}_2=2 \mathrm{~V}_1\)

\(\frac{\mathrm{nRT}_2}{\mathrm{P}_2}=\frac{2 \mathrm{nRT}_1}{\mathrm{P}_1}\)

\(\mathrm{P}_1=5, \mathrm{~T}_1=298\)

\(\mathrm{P}_2=\frac{5 \mathrm{~T}_2}{2 \times 298} \)

\(\mathrm{n} \frac{5}{2} \mathrm{R}\left(\mathrm{T}_2-\mathrm{T}_1\right)=-1\left(\frac{\mathrm{nRT}_2}{\mathrm{P}_1}-\frac{\mathrm{nRT}_1}{\mathrm{P}_1}\right) \)

Put \(\mathrm{T}_1=298\)

and \(\mathrm{P}_2=\frac{5 \mathrm{~T}_2}{2 \times 298}\)

Solve and we get \(T_2\) = 274.16 K

\(T_2 \approx 274 K\)