Question

Column I contains a list of processes involving expansion of an ideal gas. Match this with column II, describing the thermodynamic change during this process.

• A. $A\to q;B\to p,s;C\to p,r;D\to q,s$
• B. $A\to q;B\to p,r;C\to p,s;D\to q,s$
• C. $A\to q;B\to q,r;C\to p,s;D\to p,s$
• D. $A\to q;B\to p,s;C\to p,r;D\to q,r$

Answer 1

The correct option is B $A\to q;B\to p,r;C\to p,s;D\to q,s$

$A:$
As boundary is non conducting, $\Delta Q=0$. In the case of free expansion W = 0.
From the first law of thermodynamics $\Delta Q=\Delta U+\Delta W$
$0=\Delta U+0$ or $\Delta U=0$ or U is constant $\Rightarrow$ T is constant
(A) $\to \left(q\right)$ (As temperature remains constant)

$B:$
nRTV = C or TV = C'
Since when volume increases, the temperature decreases.
Now $Q=nC\Delta T$, for polytropic process,
$P{V}^x$ = constant,
$C=C_v+\frac{R}{1-x}$
$C=C_v+\frac{R}{-2+1}=C_v-R=\frac{3}{2}R-R$
or $C=\frac{R}{2}\Rightarrow Q=n\frac{R}{2}\Delta T$
$\Delta T$ is negative so Q is negative. Means heat is lost.
$\left(B\right)\to \left(p\right),\left(r\right)$

$C:$
$P{V}^{4/3}=C,T{V}^{1/3}=C$
So when volume increases temperature decreases.
Now
$C=C_v+\frac{R}{-\frac{4}{3}+1}=\frac{3}{2}R-3R$ or $C=-\frac{3}{2}R$
Hence, $Q=n(-\frac{3}{2}R)\left(\Delta T\right)$
As $\Delta T$ is negative Q will be positive
$\left(C\right)\to \left(p\right),\left(s\right)$

$D:$
As product of P and V increases, temperature increases because
$T=\frac{PV}{nR}$

$Q=\Delta U+W$
$\Delta U=+ve(\Delta T=+ve)$
$W=+ve$ (since volume increases)
so $Q=+ve$
Hence the gas gains heat,
$\left(D\right)\to \left(q\right),\left(s\right)$