One mole of a van der Waals gas obeying the equation $(\begin{matrix} P + \frac{a}{V^{2}} \end{matrix}) (V - B) = R T$ undergoes the quasi-static cyclic process which is shown in the $P - V$ diagram. The net heat absorbed by the gas in this process is

\frac{1}{2} (P_{1} - P_{2}) (V_{1} - V_{2})

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\frac{1}{2} (P_{1} + P_{2}) (V_{1} - V_{2})

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\frac{1}{2} (\begin{matrix} P_{1} + \frac{a}{v_{1}^{2}} - P_{2} - \frac{a}{v_{2}^{2}} \end{matrix}) (V_{1} - V_{2})

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\frac{1}{2} (\begin{matrix} P_{1} + \frac{a}{v_{1}^{2}} + P_{2} + \frac{a}{v_{2}^{2}} \end{matrix}) (V_{1} - V_{2})

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Solution

The correct option is A $\frac{1}{2} (P_{1} - P_{2}) (V_{1} - V_{2})$
For a cyclic process, the heat absorbed by the gas is equal to the work done in one complete cycle.
Work done in a cyclic process is equal to the area enclosed by the PV curve.
Hence net heat absorbed= $\frac{1}{2} \times B a s e \times H e i g h t$
$= \frac{1}{2} \times (V_{1} - V_{2}) \times (P_{1} - P_{2})$
$= \frac{1}{2} (P_{1} - P_{2}) (V_{1} - V_{2})$

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Similar questions

(p + \frac{a}{V^{2}}) (V - b) = R T

v_{1} + v_{2} = 4

v_{1}^{2} + v_{2}^{2} = 16

v_{1}

v_{2}

1 \to 2

P_{1} = 2 P_{2} = 10^{6} {N/ m}^{2}, V_{2} = 4 V_{1} = 0.4 m^{3}

P_{1}

V_{1}

V_{2}

V_{2}

η = \frac{γ - 1}{\frac{γ P_{2}}{P_{2} - P_{1}} + \frac{V}{V_{2} - V_{1}}}

Take P_{1} = 1 bar, P_{2} = 2 bar, v_{1} = 1 m^{3}, v_{2} = 2 m^{3}

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