ThermodynamicsHard

Question

One mole of an ideal gas undergoes the following cyclic process.

(i) Isochoric heating from (P1, V1, T1) to double temperature.

(ii) Isobaric expansion to double volume. (iii) Linear expansion (on P–V curve) to (P1, 8V1).

(iv) Isobaric compression to initial state.

If T1 = 300 K, then the magnitude of net work done by the gas in the cyclic process is

Options

A.2400 cal
B.1200 cal
C.4800 cal
D.3600 cal

Solution

$w = - \left( P_{1}V_{1} + \frac{1}{2} \times P_{1} \times 6V_{1} \right) = - 4P_{1}V_{1}$

$= - 4nRT_{1} = - 4 \times 1 \times 2 \times 300\text{ cal} $$$= - 2400\text{ cal}$$

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