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Q.(a) Write the equation of an isothermal process.

(1)
(b) Obtain an expression for work done in an isothermal process.
(3)
(c) A Carnot engine operates between the temperatures 398 K and 293 K. Find its efficiency. (1)
Kerala DhseKerala DHSE Plus One Board 2022Subjective· 5mImportance★★★★★
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An isothermal process is one carried out at constant temperature. Since the ideal gas equation is PV = nRT, holding T (and the amount of gas n) fixed makes the right-hand side a constant, so pressure and volume must vary inversely — this is Boyle's law, and it is the defining equation of an isothermal process.

Starting point — the ideal gas equation

For n moles of an ideal gas,

PV=nRTPV = nRT

where R is the universal gas constant.

Applying the isothermal condition

In an isothermal process, temperature T is held constant throughout, and the amount of gas n does not change. Since n, R, and T are all constant, the product nRT is a fixed number. Therefore

PV=constantPV = \text{constant}

So for two states (1) and (2) of the same isothermal process,

P1V1=P2V2P_1 V_1 = P_2 V_2

This is Boyle's law, and it is exactly the equation of an isothermal process — on a P–V diagram it traces out a hyperbola (an isotherm).

Work done in an isothermal process (often asked alongside this equation)

For a reversible isothermal expansion/compression from volume V₁ to V₂,

W=∫V1V2P dV=∫V1V2nRTV dV=nRTln⁡ ⁣(V2V1)=2.303 nRTlog⁡10 ⁣(V2V1)W = \int_{V_1}^{V_2} P\,dV = \int_{V_1}^{V_2} \frac{nRT}{V}\,dV = nRT\ln\!\left(\frac{V_2}{V_1}\right) = 2.303\, nRT \log_{10}\!\left(\frac{V_2}{V_1}\right)

Since internal energy of an ideal gas depends only on temperature, ΔU = 0 for any isothermal process; by the first law of thermodynamics (Q = ΔU + W), all the heat absorbed goes entirely into the work done: Q = W.

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