QJEEAdv08Phd7c6: JEE Advanced 2008 Physics - Class 11 JEEAdv08Phd7c6 Solution: An ideal gas is expanding such that PT 2 = constant. The…

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.

$$\frac{d\mathrm{V}}{\mathrm{V}}=3\frac{d\mathrm{T}}{\mathrm{T}}$$

$$\Rightarrow dV=\left(\frac{3}{T}\right)VdT$$ ..... (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,

$$VydT=\left(\frac{3}{T}\right)VdT$$

$$\Rightarrow y=\frac{3}{T}$$