Question

2 moles of an ideal diatomic gas is subjected to the following process. Mark the correct option(s).
[Given : $R=2Cal/molK$]

• A. $W=0$
• B. $q=4480Cal$
• C. $\Delta H=4480Cal$
• D. $\Delta E=1600Cal$

Answer 1

Answer: (A), (C)

$W=0$
$\Delta H=4480Cal$

Applying ideal gas equation at each of the point on the graph,
$P_1{V}_1=R{T}_1$
$P_2{V}_2=R{T}_2$

$\therefore \frac{V_1}{V_2}=\frac{\left(\frac{T_1}{P_1}\right)}{\left(\frac{T_2}{P_2}\right)}=\frac{47+273}{4}\times \frac{8}{367+273}=1$
$\therefore V_1=V_2$
$\Rightarrow \Delta V=0$
$\Rightarrow W={\int }_{V_1}^{V_2}PdV=0$ $\left(OptionA\right)$

For a diatomic gas $\gamma =\frac{f+2}{f}=\frac{5+2}{5}=\frac{7}{5}$

$C_p=\gamma C_v=\frac{\gamma R}{\gamma -1}=\frac{\frac{7R}{5}}{\frac{7}{5}-1}=\frac{7}{2}R$

$\Delta H=n{C}_p\Delta T=2\times \frac{7}{2}R\times (367-47)=2\times \frac{7}{2}\times 2\times 320=4480cal$ $\left(OptionC\right)$

$\Delta U=n{C}_v\Delta T=2\times \frac{R}{\gamma -1}\times (367-47)=2\times 2\times \frac{5}{2}\times 320=3200cal$

Since $W=0$, $\Delta Q=\Delta U=3200cal$