JEE MAIN - Physics (2003 - No. 37)
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
$${4 \over 3}$$
$$2$$
$${5 \over 3}$$
$${3 \over 2}$$
Explicació
$$P \propto {T^3} \Rightarrow P{T^{ - 3}} = $$ constant ....$$(i)$$
But for an adiabatic process, the pressure temperature relationship is given by
$${P^{1 - \gamma }}\,\,{T^\gamma } = $$ constant $$ \Rightarrow P{T^{{\gamma \over {1 - \gamma }}}} = $$ constant. ....$$(ii)$$
From $$(i)$$ and $$(ii)$$ $${\gamma \over {1 - \gamma }} = - 3 \Rightarrow \gamma = - 3 + 3\gamma \Rightarrow \gamma = {3 \over 2}$$
But for an adiabatic process, the pressure temperature relationship is given by
$${P^{1 - \gamma }}\,\,{T^\gamma } = $$ constant $$ \Rightarrow P{T^{{\gamma \over {1 - \gamma }}}} = $$ constant. ....$$(ii)$$
From $$(i)$$ and $$(ii)$$ $${\gamma \over {1 - \gamma }} = - 3 \Rightarrow \gamma = - 3 + 3\gamma \Rightarrow \gamma = {3 \over 2}$$