JEE MAIN - Physics (2023 - 30th January Morning Shift - No. 5)
The pressure $$(\mathrm{P})$$ and temperature ($$\mathrm{T})$$ relationship of an ideal gas obeys the equation
$$\mathrm{PT}^{2}=$$ constant. The volume expansion coefficient of the gas will be :
$$3 T^{2}$$
$$\frac{3}{T^2}$$
$$\frac{3}{T^3}$$
$$\frac{3}{T}$$
Explanation
$$PT^2$$ = constant
From $$PV = nRT \Rightarrow {{{T^3}} \over V} = $$ constant
$${T^3} \propto V$$ ..... (1)
$$3{T^2}dT \propto dV$$ ..... (2)
From (1) and (2)
$${{3dT} \over T} = {{dV} \over V}$$
$$\therefore$$ $$\gamma = {1 \over V}{{dV} \over {dT}} = {3 \over T}$$
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