# In thermodynamics, the 𝑃 - 𝑉 work done is given by

$$w = -\int dVP_{ext}$$. For a system undergoing a particular process, the work done is,

$$w = -\int dV \left (\frac{RT}{V-b}-\frac{a}{V^{2}} \right )$$

This equation is applicable to a

a) System that satisfies the van der Waals equation of state.

b) Process that is reversible and isothermal.

c) Process that is reversible and adiabatic.

d) Process that is irreversible and at constant pressure.

Answer: a, b, c

$$w = -\int P_{ext.} dV$$

From van der waal equation of state for one mole gas.

$$\left ( P+\frac{a}{V^{2}} \right )(v-b) = RT$$ $$P = \left [ \frac{RT}{V-b}-\frac{a}{V^{2}} \right ]$$

For reversible process

Pext. = Pgas

$$W = -\int \left ( \frac{RT}{V-b}-\frac{a}{V^{2}} \right )dv$$

So, process is not applicable only for irreversible process.

Hence Ans, a, b, c

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