Question

One mole of an ideal gas is subjected to a two-step reversible process (A - B and B - C). The pressure at A and C is the same. Mark the correct statement(s):

• A. Work done in the path AB is zero
• B. In the path AB, work will be done on the gas by the surrounding
• C. $V_C=3\times V_A$
• D. $V_B=16.42\text{litres}$

Answer 1

Answer: (C), (D)

The correct options are
C $V_C=3\times V_A$
D $V_B=16.42\text{litres}$

$A\to B:$ Straight line but not through origin. Hence, 'V' is not constant. So ${}^{\prime }{\omega }^{\prime }.$ is not zero.
now, $\frac{P}{T}=\frac{nR}{V}or\frac{T}{P}=\frac{V}{nR}\therefore V\alpha \frac{T}{P}$
at $A:\frac{T}{P}=\frac{100}{1}=100$
at $B:\frac{T}{P}=\frac{600}{3}=200$
$\therefore V_B>V_A\Rightarrow$ Expansion.
$\therefore$ work is done by the gas from A to B.
$P_C=P_A\therefore \frac{V_C}{T_C}=\frac{V_A}{T_A}\Rightarrow V_C=\frac{T_C}{T_A}.V_A$
or $V_C=\frac{300}{100}\times V_A=3V_A$
$V_B=\frac{nR{T}_B}{P_B}=\frac{1\times 0.0821\times 600}{3}=16.42L$