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 :
Solution
<p>$PT^2$ = constant</p>
<p>From $PV = nRT \Rightarrow {{{T^3}} \over V} =$ constant</p>
<p>${T^3} \propto V$ ..... (1)</p>
<p>$3{T^2}dT \propto dV$ ..... (2)</p>
<p>From (1) and (2)</p>
<p>${{3dT} \over T} = {{dV} \over V}$</p>
<p>$\therefore$ $\gamma = {1 \over V}{{dV} \over {dT}} = {3 \over T}$</p>
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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