11.Thermodynamics
normal

During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio ${C_p}/{C_v}$ for the gas is

A

$1.5$

B

$1.33$

C

$2$

D

$1.67$

Solution

(a) Given $P \propto {T^3}$, but we know for an adiabatic process, the pressure $P \propto {T^{\gamma /\gamma – 1}}$
So $\frac{\gamma }{{\gamma – 1}} = 3 \Rightarrow \gamma = \frac{3}{2} \Rightarrow \frac{{{C_P}}}{{{C_V}}} = \frac{3}{2}$

Standard 11
Physics

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