Question

One mole of a monoatomic ideal gas undergoes the process $A\to B$ , as in the given $P-V$ diagram. The specific heat for this process is

• A. $\frac{3R}{2}$
• B. $\frac{15R}{7}$
• C. $\frac{30R}{7}$
• D. $\frac{20R}{7}$

Answer 1

Answer: (B)

$\Delta Q=\Delta U+W$ (From $1^{st}$ law)
Here, $\Delta U=n{C}_V\Delta T$
Temperature at $A$ ,
$T_A=\frac{2P_0{V}_0}{R}$
At $B$ , $T_B=\frac{16P_0{V}_0}{R}$
So, $\Delta U=\frac{3}{2}R\left[\frac{14P_0{V}_0}{R}\right]=21P_0{V}_0$
$W=\left(2P_0\right)\left(3V_0\right)+\frac{1}{2}\left(3V_0\right)\left(2P_0\right)$ [Area under $P-V$ diagram]
$\Rightarrow 9P_0{V}_0$
So, $\Delta Q=21P_0{V}_0+9P_0{V}_0$
$\Rightarrow 30P_0{V}_0$
$\Delta Q=nC\Delta T$
$C=\frac{\Delta Q}{n\Delta T}=\frac{30P_0{V}_0}{\frac{14P_0{V}_0}{R}}=\frac{15}{7}R$