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Q.(a) Derive the relation between 'Cp' and 'Cv' for an ideal gas.

(2)
(b) Explain the following:
(i) Free expansion.
(1)
(ii) Isolated system. (1)
Himachal HpboseHPBOSE Himachal Pradesh Class 11 Board Exam 2024Subjective· 4mImportance★★★★★
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For an ideal gas, Cp exceeds Cv by exactly R because constant-pressure heating must also supply energy for expansion work; free expansion does zero work (into a vacuum); an isolated system exchanges neither matter nor energy with its surroundings.

(a) Derivation of Cp - Cv = R:

By definition, for one mole of gas:

  • CV=(∂U∂T)VC_V = \left(\dfrac{\partial U}{\partial T}\right)_V, so at constant volume, dU=CV dTdU = C_V \, dT (all the heat supplied at constant volume increases internal energy only, since no PV work is done when volume is fixed).
  • CP=(∂H∂T)PC_P = \left(\dfrac{\partial H}{\partial T}\right)_P, so at constant pressure, dH=CP dTdH = C_P \, dT.

Enthalpy is defined as H=U+PVH = U + PV. For one mole of an ideal gas, PV=RTPV = RT, so:

H=U+RTH = U + RT

Differentiating with respect to temperature:

dH=dU+R dTdH = dU + R\, dT

Substituting dH=CP dTdH = C_P\,dT and dU=CV dTdU = C_V\,dT:

CP dT=CV dT+R dTC_P \, dT = C_V \, dT + R \, dT

Dividing through by dTdT:

CP−CV=R\boxed{C_P - C_V = R}

This makes physical sense: heating a gas at constant pressure requires supplying extra energy (equal to R per degree per mole) beyond what's needed at constant volume, because the gas must also do expansion (PV) work on its surroundings as it expands.

(b)(i) Free expansion: …

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