Question

One mole of a monoatomic ideal gas undergoes the process $A\to B$ in the given p-V diagram. The molar heat capacity for this process is

• A. $\frac{3R}{2}$
• B. $\frac{13R}{6}$
• C. $\frac{5R}{2}$
• D. $2R$

Answer 1

Answer: (B)

$\frac{13R}{6}$

$W_{A\to B}=$ Area under p-V diagram
$=4V_0\times 3p_0+\frac{1}{2}4V_0\times 3p_0=18p_0{V}_0$
$\therefore \Delta U=n{C}_v\Delta T$
$=1\cdot \frac{3}{2}R(T_B-T_A)$
$=\frac{3}{2}R(\frac{30p_0{V}_0}{R}-\frac{3p_0{V}_0}{R})$
$=\frac{81}{2}{p}_0{V}_0$
Thus, $\Delta Q_{A\to B}=\Delta W_{A\to B}+\Delta U_{A\to B}$
$=18p_0{V}_0+\frac{81}{2}{p}_0{V}_0=\frac{117}{2}{p}_0{V}_0$
$\therefore$ Molar heat capacity, $C=\frac{\Delta Q}{n\Delta T}$
$=\frac{117p_0{V}_0/2}{\frac{30p_0{V}_0}{R}-\frac{3p_0{V}_0}{R}}=\frac{117p_0{V}_0/2}{\frac{27p_0{V}_0}{R}}=\frac{13R}{6}$