Question

The figures given below show different processes (relating pressure $P$ and volume $V$) for a given amount for an ideal gas. $\Delta W$ is work done by the gas and $\Delta Q$ is heat absorbed by the gas.

$\begin{array}{cc}\text{Column I}& \text{Column II}\\ 1.\text{In Fig}\left(i\right)& \left(p\right)\Delta Q>0\\ 2.\text{In Fig}\left(ii\right)& \left(q\right)\Delta W<0\\ 3.\text{In Fig}\left(iii\right)& \left(r\right)\Delta Q<0\\ 4.\text{In Fig}\left(iv\right)& \left(s\right)\Delta W>0\end{array}$

• A. $1\to (p,r)2\to q,3\to (p,r),4\to (p,r)$
• B. $1\to (p,r),2\to s,3\to (p,s),4\to (p,r)$
• C. $1\to (p,s),2\to s,3\to (p,s),4\to (q,r)$
• D. $1\to (p,s),2\to p,3\to (p,r),4\to (p,s)$
  

Answer 1

The correct option is C $1\to (p,s),2\to s,3\to (p,s),4\to (q,r)$

For figure $\left(i\right)$
Applying $PV=nRT$
$P=\left(nRT\right)\frac{1}{V}$
since from graph $P\propto \frac{1}{V}$
$\Rightarrow T=\text{constant}$, hence it is isothermal process
$\because (\frac{1}{V}\downarrow )$ that means $V$ increases, $\Delta W$ is positive.
and $\Delta Q=\Delta U+\Delta W=0+\Delta W$
$\Rightarrow W>0\&\Delta Q>0$

For figure $\left(ii\right)\Delta Q=0$,being an adiabatic process
As the gas is expanding hence $W>0$ and applying first law of thermodynamics,
$0=\Delta U+\Delta W$
$\Delta U=-\Delta W\Rightarrow \Delta U<0$

For figure $\left(iii\right)$applying $PV=nRT$
As volume increases, $T$ also increases i.e $U\uparrow$
$\Rightarrow \Delta U>0$
work done by gas is $+ve$ due to expansion. Applying first law of thermodynamics,
$\Delta Q=\Delta U+\Delta W$
$\Rightarrow \Delta Q>0$

For figure $\left(iv\right)$ For cyclic process $\Delta U=0$
$\&\Delta W<0$ (anticlockwise cycle)
Hence from first law of thermodynamics we get,
$\Delta Q=\Delta U+\Delta W$
$\Rightarrow \Delta Q<0$.
$\therefore$$1\to (p,s),2\to s,3\to (p,s),4\to (q,r)$ is correct.