Question

$N_2$ gas is assumed to behave ideally A given volume of $N_2$ originally at 373 k and 0.1013 M pa pressure is adiabatically compressed due to which its temperature rises to $673K(Cv=\frac{5}{2}R)$

Which of the following statement(s) is/are correct?

• A. The change in internal energy is 6235.5 J $mol{e}^{-1}$
• B. In this case the final internal pressure is equal to the external pressure
• C. The final pressure of $N_2$ is approximately 0.38 MPa
• D. The final pressure of $N_2$ is approximately 0.02 Mpa

Answer 1

Answer: (A), (B), (C)

The correct options are
A The change in internal energy is 6235.5 J $mol{e}^{-1}$
B In this case the final internal pressure is equal to the external pressure
C The final pressure of $N_2$ is approximately 0.38 MPa

$\Delta E=\int P_{ext}dV=-P_{ext}(V_2-V_1)$

$(T_2-T_1)=-P_{ext}(\frac{R{T}_2}{P_2}-\frac{R{T}_1}{P_1})$ Here, $P_{ext}=P_2$

$(T_2-T_1)=-P_2[\frac{R{T}_2}{P_2}-\frac{R{T}_1}{P_1}]$

$\Rightarrow \frac{5}{2}R(673-373)=-P_2[\frac{R\left(673\right)}{P_2}-\frac{R\left(373\right)}{0.1013}]$

$\Rightarrow \frac{5}{2}\times 300=-673+\frac{373P_2}{0.1013}$

$\therefore P_2=\frac{(750+673)}{373}\times 0.1013=0.3865MPa$

$\Delta E=Cv(T_2-T_1)=\frac{5}{2}R\times 300=\frac{5}{2}R\times 300=\frac{5}{2}\times 8.314\times 300J/mol=6235.5J/mol$

Hence, A, B and C are correct.