Isochoric Process

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Heat of 400 J is supplied to a gas sealed inside a rigid steel container whose volume cannot change, so what is the work done by the gas?

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As described in NCERT Class 11, Chapter 12 (Thermodynamics), the work done by a gas during a volume change is $\Delta W = \int P\,dV$. In an isochoric (constant volume) process the volume is fixed, so $dV = 0$ and therefore $\Delta W = 0$, no matter how much pressure or temperature rises. A rigid steel container enforces exactly this condition. By the First Law, $\Delta Q = \Delta \cup + \Delta W = \Delta \cup + 0$, so all 400 J of supplied heat goes entirely into raising the internal energy of the gas, increasing its temperature. The option that work equals 400 J is wrong because that would require the heat to be converted to mechanical work, which is impossible without volume change. The option of 200 J is an arbitrary fraction with no physical basis here. The option that it cannot be determined is wrong because the constant-volume condition alone fixes the work at zero independent of temperature. A check on units and logic: $P\,dV$ has units of energy, and with $dV = 0$ the integral is identically zero, confirming the result.

Isochoric Process

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