JEE Advance - Physics (2008 - Paper 1 Offline - No. 5)
An ideal gas is expanding such that PT$$^2$$ = constant. The coefficient of volume expansion of the gas is
$$\frac{1}{\mathrm{T}}$$
$$\frac{2}{\mathrm{T}}$$
$$\frac{3}{\mathrm{T}}$$
$$\frac{4}{\mathrm{T}}$$
Explanation
Given, PT$$^2$$ = constant ..... (i)
For an ideal gas,
$$\frac{\mathrm{PV}}{\mathrm{T}}$$ = constant ...... (ii)
From above two equation, after eliminating P.
$$\frac{\mathrm{V}}{\mathrm{T^3}}$$ = Constant
$$\Rightarrow$$ V = KT$$^3$$, where $$k$$ = constant.
$$\frac{d\mathrm{V}}{\mathrm{V}}=3\frac{d\mathrm{T}}{\mathrm{T}}$$
$$ \Rightarrow dV = \left( {{3 \over T}} \right)VdT$$ ..... (iii)
Change in volume due to thermal expansion is given by $$dV=Vydt$$ ..... (iv)
Where, $$y$$ = coefficient of volume expansion
From equation (iii) and (iv), we have,
$$VydT = \left( {{3 \over T}} \right)VdT$$
$$ \Rightarrow y = {3 \over T}$$
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