Question

An ideal monoatomic gas undergoes different types of processes which are described in column I. Match the corresponding statements in column II. The parameters involved have standard usual meaning.
$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \left(A\right)P=2V^2& \text{(p) If volume increases then}\\ & \text{temperature will also increase}\\ \left(B\right)P{V}^2=\text{constant}& \text{(q) If volume increases then}\\ & \text{temperature will decrease}\\ \left(C\right)C=C_V+2R& \text{(r) For expansion, heat will have}\\ & \text{to be supplied to gas}\\ \left(D\right)C=C_V-2R& \text{(s) If temperature increases then the}\\ & \text{workdone by gas is positive}\\ & \text{(t) If temperature decreases then workdone by gas is}\\ & \text{positive}.\end{array}$

• A. $A\to p,r,sB\to q,tC\to p,r,sD\to q,r,t$
• B. $A\to p,tB\to q,rC\to q,sD\to q,r,s$
• C. $A\to q,tB\to p,r,sC\to p,r,sD\to q,r,s$
• D. $A\to q,r,sB\to q,tC\to p,r,sD\to p,r,t$

Answer 1

The correct option is A $A\to p,r,sB\to q,tC\to p,r,sD\to q,r,t$

If $P=2V^2$, then using ideal gas equation, $PV=nRT$
$\Rightarrow \left(2V^2\right)\left(V\right)=nRT$
or, $nRT=2V^3$
or, $T\propto V^3$
Hence if volume $\uparrow$, temperature of gas will also increase.

Increase in temperature means $dT$ is +ve.
or, $dU\propto dT$
$\therefore dU$ is also +ve.

Using first law of thermodynamics;
$dQ=dU+dW$
$\Rightarrow dQ$ is also +ve
$(\because dW$ is +ve for expansion of gas $(V\uparrow ))$
Hence heat is supplied during expansion of gas.
$\therefore \left(A\right)\to p,r,s$



(B) If $P{V}^2=\text{constant}$, using ideal gas equation $PV=nRT$
$\Rightarrow \left(nRT\right)V=\text{constant}$
$VT=\text{constant}$
$V\propto \frac{1}{T}$
Hence if volume of gas increases, its temperature will decrease.

For polytropic process,
$C=C_V+\frac{R}{1-x}$
$C=\frac{3R}{2}-R=\frac{R}{2}$

Heat supplied is $dQ=nCdT$
If temperature decreases $dQ$ is negative


Now, for rise in temperature $(dT\to +ve)$ the volume decrease $(V\downarrow )$ or workdone by gas is -ve. However the heat supplied to gas in this case.
$dQ\propto dT$
$dT\to +ve$ means $dQ$ is also +ve.
$\Rightarrow \left(B\right)\to q,t$


(C) Given, $C=C_V+2R$
For polytropic process,
$C=C_V+\frac{R}{1-x}$
$1-x=0.5$
$x=0.5$

$P\sqrt{V}=\text{Constant}$
Using $PV=nRT$
$\frac{T}{\sqrt{V}}=\text{constant}$

Hence with increase in temperature, $V\uparrow$
Also if $V\uparrow$ then $dW$ is +ve and $dU$ is +ve $(T\uparrow )$
From relation (i), $dQ=dU+dW$
$dQ$is +ve, if $dU$ & $dW$ are positive (when temperature of gas increases)

Hence heat is supplied $(dQ\to +ve)$ when gas is expanding $\left[\right(V\uparrow )$ or $dW$ is +ve]
$\Rightarrow \left(C\right)\to p,r,s$


(D) Given, $C_V-2R=C$
$C=C_V+\frac{R}{1-x}$
$1-x=-0.5$
$x=1.5$

$P{V}^{1.5}=\text{Constant}$
$T\sqrt{V}=\text{Constant}$

Hence with increase in temperature, the volume of gas decreases.
$\Rightarrow T\uparrow (dT$ is +ve) then $V\downarrow$ (compression of gas), $dW$ is -ve.

Also, $dQ=n(C_V-2R)dT$
For monoatomic gas $C_V=\frac{3}{2}R$ i.e, $C<0$

Thus $dQ$ is -ve when temperature increases, or we can say heat is supplied to gas $(dQ\to +ve)$ when its temperature decrease $(T\uparrow )$ or volume increases.
$\therefore \left(D\right)\to q,r,t$
$\begin{array}{c}\text{Why this question}?\\ \text{Tip: It tests your analytical skills in relating}\\ \text{the workdone by gas, change in temperature}\&\\ \text{volume by correlating the ideal gas equation}\\ \text{(PV =nRT) with specific heat capacity of gas}\\ \text{and first law of thermodynamics (dQ = dU + dW)}\end{array}$