Question

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

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

Answer 1

Answer: (B)

$\Delta Q=\Delta U+W$ (From $1^{s t}$ law)
Here, $\Delta U=nC_{V}\Delta T$
Temperature at $A$ ,
$T_{A}=\frac{2 P_{0} V_{0}}{R}$
At $B$ , $T_{B}=\frac{16 P_{0} V_{0}}{R}$
So, $\Delta U=\frac{3}{2}R\left[\frac{14 P_{0} V_{0}}{R}\right]=21P_{0}V_{0}$
$W=\left(2 P_{0}\right)\left(3 V_{0}\right)+\frac{1}{2}\left(3 V_{0}\right)\left(2 P_{0}\right)$ [Area under $P-V$ diagram]
$\Rightarrow 9P_{0}V_{0}$
So, $\Delta Q=21P_{0}V_{0}+9P_{0}V_{0}$
$\Rightarrow 30P_{0}V_{0}$
$\Delta Q=nC\Delta T$
$C=\frac{\Delta Q}{n \Delta T}=\frac{30 P_{0} V_{0}}{\frac{14 P_{0} V_{0}}{R}}=\frac{15}{7}R$