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.

Column 1	Column 2
i. In Fig. (i)	(a). $\Delta Q>0$
ii. In Fig. (ii)	(b). $\Delta W<0$
iii. In Fig. (iii)	(c). $\Delta Q<0$
iv. In Fig. (iv)	(d). $\Delta W>0$

• A. $i\to c;ii\to b;iii\to b,c;iv\to b$
• B. $i\to b,c;ii\to d;iii\to c,b;iv\to d$
• C. $i\to c,b;ii\to a,d;iii\to b;iv\to a$
• D. $i\to a,d;ii\to d;iii\to a,d;iv\to b,c$

Answer 1

The correct option is D $i\to a,d;ii\to d;iii\to a,d;iv\to b,c$

(a) $PV=nRT$
$P=\left(nRT\right)\frac{1}{V}=(const.)\frac{1}{V}$
$T=$constant i.e. isothermal process $(\Delta U=0)$
$\therefore$ $V$ increases. $pdV=\Delta WasdVispositive,\Delta W>0\Delta Q=\Delta U+\Delta W\Delta Q=\Delta Was\Delta Wispositive,\Delta Q>0$

(b). In adiabatic process$\Delta Q=0pdV=\Delta W=positive$

(c). $PV=nRT$
As volume increases, $T$ also increasse
i.e. $\Delta U>0pdV=\Delta W>0,So\Delta Q>0$

(d).For cyclic process $\Delta U=0$
$\Delta W<0\left(anticlockwise\right)$
$\Delta Q<0$