A) \[\frac{\gamma +1}{\gamma -1}\]
B) \[\frac{\gamma -1}{\gamma +1}\]
C) \[\frac{1}{2}(\gamma -1)\]
D) \[\frac{2}{\gamma -1}\]
Correct Answer: D
Solution :
Let us consider 1 mole of an ideal gas at absolute temperature T. It has N molecules. In an ideal gas there are no intermolecular forces i.e., there is no internal potential energy. The internal energy of an ideal gas is entirely kinetic energy of its molecules. Average kinetic energy per molecule of an ideal gas is \[\frac{1}{2}fkT,\]where/is number of degrees of freedom. The internal energy of 1 mole of an ideal gas is \[U=N\times \frac{1}{2}fkT=\frac{1}{2}fRT\] \[\left( \because \,k\frac{R}{N} \right)\] Differentiating it, we get \[\frac{dU}{dT}=\frac{1}{2}fR\] Let the gas is heated at constant volume until its temperature rises through dT. Heat given is \[dQ={{C}_{v}}dT,\]and no external work is done. From law of thermodynamics \[dU=dQdW\] We have \[dU={{C}_{v}}dT\] \[{{C}_{V}}=\frac{dU}{dT}\] Putting the value of \[\frac{dU}{dT},\] we have \[{{C}_{V}}=\frac{1}{2}fR\] This is molar specific heat of gas at constant volume. From Mayor's formula \[{{C}_{P}}-{{C}_{V}}=R\] \[\therefore \] \[{{C}_{P}}={{C}_{v}}+R=\frac{1}{2}fR+R=\left( \frac{f}{2}+1 \right)R\] The ratio \[\gamma =\frac{{{C}_{P}}}{{{C}_{V}}}=\frac{\left( \frac{f}{2}+1 \right)R}{\frac{f}{2}R}\] \[\Rightarrow \] \[f=\frac{2}{\gamma -1}\]
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