InterviewSolution
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InterviewSolution
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A gas undergoes a process such that `pprop1/T`. If the molar heat capacity for this process is `C=33.24J//mol-K`, find the degree of freedom of the molecules of the gas. |
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Answer» Correct Answer - D As `pprop1/T` or `pT=` constant …(i) We have for one mole of an ideal gas `pV=RT` …(ii) From Eqs. (i) and (ii), `p^2V=` constant or `pV^(1//2)=K (say) =p_iV_i^(1//2)=p_fV_f^(1//2)` …(iii) From first law of thermodynamics, `DeltaQ=DeltaU+DeltaW` or `CDeltaT=C_VDeltaT+DeltaW` or `C=C_V+(DeltaW)/(DeltaT)` ...(iv) Here, `DeltaW=intpdV=Kint_(V_i)^(V_f)V^(-1//2)dV` `=2K[V_f^(1//2)-V_i^(1//2)]=2[p_fV_f^(1//2)V_f^(1//2)-p_iV_i^(1//2)V_i^(1//2)]` `=2[p_fV_f-p_iV_i]=2R[T_f-T_i]` `=(RDeltaT)/(1//2)implies(DeltaW)/(DeltaT)=2R` Substituting in Eq. (iv), we have `C=C_V+2R=(R)/(gamma-1)+2R` Substituting the value, `33.24=R((1)/(gamma-1)+2)=8.31((1)/(gamma-1)+2)` Solving this we get `gamma=1.5` Now, `gamma=1+2/F` or degree of freedom `F=(2)/(gamma-1)=(2)/(1.5-1)=4` |
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