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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