A) \[{{C}_{p}}-{{C}_{v}}=R\]
B) \[{{C}_{p}}-{{C}_{v}}<R\]
C) \[{{C}_{p}}-{{C}_{v}}>R\]
D) \[{{C}_{p}}-{{C}_{v}}=0\]
Correct Answer: A
Solution :
Key Idea: The difference in \[{{\text{C}}_{\text{P}}}\] and \[{{\text{C}}_{\text{V}}}\]is equal to that amount of work which is done against the external pressure to raise the temperature of 1 mole of gas through \[\text{1}{{\,}^{\text{o}}}\text{C}\] Let the volume of 1 mole of gas at temperature T be \[{{\text{V}}_{\text{1}}}\] and at temperature \[\text{(T}\,\text{+}\,\text{1)}\]be \[{{\text{V}}_{2}}.\]The pressure P remains constant in both the states. Then, external work = pressure \[\times \] change in volume \[{{C}_{P}}-{{C}_{V}}=P\times ({{V}_{2}}-{{V}_{1}})\] From gas equation, \[P{{V}_{1}}=RT,P{{V}_{2}}=R(T+1)\] Subtracting \[P({{V}_{2}}-{{V}_{1}})=R(T+1)-RT=R\] \[\therefore \] \[{{C}_{P}}-{{C}_{V}}=R\] Note: This equation \[{{C}_{P}}-{{C}_{V}}=R\]is called Mayors formula. In this formula the values of \[{{C}_{P}},{{C}_{V}}\]and R should be in the same unit. All the three should be either in \[J/mol.{{\,}^{o}}C\] or in \[cal/mol\,{{.}^{o}}C.\]
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