PYQ NEET- Chemical Thermodynamics L-5

Question: A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of $2.50 \mathrm{~L}$ to a final volume of $4.50 \mathrm{~L}$. The change in internal energy $\Delta U$ of the gas in joules will be#

A) $1136.25 \mathrm{~J}$

B) $-500 \mathrm{~J}$

C) $-505 \mathrm{~J}$

D) $+505 \mathrm{~J}$

Answer: $-505 \mathrm{~J}$#

Solution:#

Key concept According to first law of thermodynamics, $$ \Delta U=q+w $$ where, $\Delta U=$ internal energy $q=$ heat absorbed or evolved, $w=$ work done.

Also, work done against constant external pressure (irreversible process). $$ w=-p_{\text {ext }} \Delta V . $$

Work done in irreversible process, $$ \begin{aligned} w & =-p_{\text {ext }} \Delta V=-p_{\text {ext }}\left(V_2-V_1\right) \ & =-2.5 \mathrm{~atm}(4.5 \mathrm{~L}-2.5 \mathrm{~L}) \ & =-5 \mathrm{~L} \operatorname{atm}=-5 \times 101.3 \mathrm{~J} \ & =-505 \mathrm{~J} \end{aligned} $$

Since, the system is well insulated, $q=0$ $$ \therefore \quad \Delta U=w=-505 \mathrm{~J} $$

Hence, change in internal energy, $\Delta U$ of the gas is $-505 \mathrm{~J}$.