Chemistry - Thermodynamics - I Question with Solution | TestHub

Given:

$E=K{V}^2$

For a monoatomic ideal gas:

$E=\Delta U$

Differentiate:

$dU=2KVdV$

Using the first law of thermodynamics:

$dU=q+w$

For a reversible process:

$w=-PdV$

From the ideal gas law:

$U=\frac{3}{2}PV=K{V}^2$

$P=\frac{2K}{3}V$

So,

$w=-\frac{2K}{3}VdV$

$\Delta U=2KVdV$

$w=-\frac{\Delta U}{3}$

$q=\Delta U-w=\frac{4}{3}\Delta U$

Also,

$P\propto V$

$T\propto PV\propto V^2$

$P\propto \sqrt{T}$

Therefore, all options are correct:

A, B, C, D