First slide

Heat ;work and internal energy

Question

If $γ$ is the ratio of two specific heats of a gas, the number of degrees of freedom of a molecule of the gas is:

Moderate

A

$\frac{25}{2} (γ-1)$
B

$\frac{3 γ - 1}{2 γ - 1}$
C

$\frac{2}{γ - 1}$
D

$\frac{9}{2} (γ-1)$

Solution

$\begin{array}{rcl} {molar specific heat of a gas at constant volume is C}_{v} & = & f. \frac{R}{2} \\ h e r e f i s d e g r e e o f f r e e d o m, \\ R i s u n i v e r s a l g a s c o n s \tan t \\ f r o m d i f f e r e n c e o f s p e c i f i c h e a t s \\ C_{p} & = & C_{v} + R \\ s u b s t i t u t e t h e v a l u e o f C_{v} \\ = & \frac{fR}{2} +R =fR (\frac{1}{2} + \frac{1}{f}) \\ C_{p} =fR (\frac{f+2}{2f}) \\ s p e c i f i c h e a t r a t i o γ & = & \frac{C_{p}}{C_{v}}, s u b s t i t u t e t h e v a l u e s o f s p e c i f i c h e a t s \\ = \frac{fR (\frac{f+2}{2f})}{\frac{fR}{2}} \\ \Rightarrow & γ = \frac{f+2}{f} \\ \Rightarrow & γ = 1 + \frac{2}{f} \\ \Rightarrow & \frac{2}{f} = γ-1 \\ \Rightarrow & d e g r e e o f f r e e d o m i s f = \frac{2}{γ - 1} \end{array}$

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