Explanation

(d) Consider $n$ moles of a gas which undergo isochoric process, $ie$, $V$=constant. From first law of thermodynamics,
$\mathcal{\mathrm{\Delta}Q = \mathrm{\Delta}}W + \mathrm{\Delta}U$ …(i)
Here, $\mathrm{\Delta}W = 0\ as\ V = constant$
$\mathcal{\mathrm{\Delta}Q =}nC_{V}\mathrm{\Delta}T$
Substituting in Eq. (i), we get
$\mathrm{\Delta}U = nC_{V}\mathrm{\Delta}T$ …(ii)
Mayer’s relation can be written as
$C_{p} - C_{V} = R$
$\Rightarrow$ $C_{V} = C_{p} - R$ …(iii)
From Eqs. (ii) and (iii), we have
$\mathrm{\Delta}U = n(C_{p} - R)\mathrm{\Delta}T$
Given, $n = 6,\ \ \ C_{p} = 8\ cal\ mol^{- 1} - K^{- 1}$,
$R = 8.31\ J\ mol^{- 1} - K^{- 1}$
$\approx 2\ cal\ mol^{- 1} - K^{- 1}$
Hence, $\mathrm{\Delta}U = 6(8 - 2)(35 - 20)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 6 \times 6 \times 15 = 540\ cal$