Question

One mole of an ideal monatomic gas is taken from temperature $T_0$ to $2T_0$ by the process $T^4{P}^{-1}=C$. Then,

• A. molar specific heat of the gas is $-\frac{3R}{2}$
• B. molar specific heat of the gas is $\frac{3R}{2}$
• C. work done by the gas is $-3R{T}_0$
• D. heat released by the gas is $\frac{3}{2}R{T}_0$

Answer 1

Answer: (A), (C), (D)

The correct options are
A molar specific heat of the gas is $-\frac{3R}{2}$
C work done by the gas is $-3R{T}_0$
D heat released by the gas is $\frac{3}{2}R{T}_0$

Given $T^4{P}^{-1}=constant$
Substituting $T=\frac{PV}{R}$, we get $P^3{V}^4=const$
$\therefore P{V}^{\frac{4}{3}}=const$
For a polytropic process $P{V}^{\mu }=constant$ where $\mu$ is the polytropic exponent
$\Rightarrow \mu =\frac{4}{3}$ by comparison.
For a monoatomic gas, $\gamma =\frac{5}{3}$
Molar specific heat:
$c=\frac{R}{\gamma -1}-\frac{R}{\mu -1}=\frac{R}{\frac{5}{3}-1}-\frac{R}{\mu -1}=\frac{3R}{2}-3R=-\frac{3R}{2}$
Heat absorbed by the system: $Q=c\times (2T_0-T_0)=-\frac{3R}{2}{T}_0$
Since $Q$ is negative, heat is released by the gas.
Work done by the gas: $W={\int }_a^{b}PdV=R\frac{T_1-T_2}{\mu -1}=R\frac{T_0-2T_0}{\frac{4}{3}-1}=-3R{T}_0$