Question

A gas occupies $2L$ at S.T.P. It is provided with $300Joule$ of heat so that its volume becomes $2.5litre$ at a pressure of $1atm$. The value of $\Delta U$(Change in internal energy) of process is :-

• A. $350.5Joule$
• B. $249.5Joule$
• C. $150.35Joule$
• D. None of these

Answer 1

The correct option is B $249.5Joule$

Given:

Initial volume of gas,$V_i=2l$

Heat provided, Q= 300J

Final volume, $V_f=2.5l$

Final pressure,$P_f=1atm$

Solution:

We know that the work is done at constant pressure and thus is irreversible.

Work done, W=-P(Change in volume)

$\Rightarrow W=-1\times (2.5-2)$

$\Rightarrow W=-0.5atm-l$

$\Rightarrow W=-0.5\times 101325\times \frac{1}{1000}J$

$\Rightarrow W=-50.6625J$

From the first law of thermodynamics,

$Q=\delta U-W$, where$\delta U=Changeininternalenergy$

$\Rightarrow \delta U=Q+W$

$\Rightarrow \delta U=300+(-50.6625)$

Therfore, $\delta U = 249.3Joule$

Hence, correct option is B.