Question 28: Thermodynamics & KTG

Question

One mole of an ideal diatomic gas undergoes a process as shown in the figure. The molar specific heat of the gas in the process is
![](https://app-content.cdn.examgoal.net/fly/@width/image/1lhyea6gr/48bc488b-d287-48ea-ab8b-9ae44dfa1ef6/cc5fb7c0-f85f-11ed-a585-812e165f6969/file-1lhyea6gs.png?format=png)

Accepted Answer

In the curve, V $\propto$ $\frac{1}{T}$
$\Rightarrow$ V = m$\frac{1}{T}$ where, m = slope of curve.
$\Rightarrow$ VT = constant
$	herefore$ TdV + VdT = 0 ..... (i)
$\Rightarrow$ dV = $-\frac{V}{T}$dT
As, Q = $\Delta$U + W
CdT = C
V
dT + pdV ..... (ii)
CdT = C
V
dT $-$ $\frac{p}{T}$ VdT
$\Rightarrow$ C = C
V
$-$ $\frac{pV}{T}$
= C
V
$-$ R ($\because$ $\frac{pV}{T}$ = R)
For diatomic gas, C
V
= $\frac{5}{2}$R
$	herefore$ C = C
V
$-$ R = $\frac{5}{2}$R $-$ R = $\frac{3R}{2}$

Answer

3R/2

Answer

R/2

Answer

5R/2

Answer

7R/2

Thermodynamics & KTG

One mole of an ideal diatomic gas undergoes a process as shown in the figure. The molar specific heat of the gas in the process is

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