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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