Question 24: 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

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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