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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 same. Mark the correct
statement(s):

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Volume of gas at B is 16.42 litres

Work involved in the path AB is zero

Volume of gas at C =3 × volume of gas at A

In the path AB work will be done on the gas by the surrounding

answer is C, D.

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Detailed Solution

$\begin{aligned} At, V_{A} = \frac{1 \times R \times 100}{1} = 100 R \\ V_{B} = \frac{1 \times R \times 600}{3} = 200 R \end{aligned}$

$∵ V_{B} > V_{A}$

so expansion of gas takes place

$V_{B} = 200 \times 0 .0821 = 16 .42 L$

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