Question

What is the change in internal energy when a gas contracts from $377mL$ to $177mL$ under a constant pressure of $1520$ torr, while at the same time being cooled by removing $124J$ heat?

• A. $40.52J$
• B. $-83.48J$
• C. $-248J$
• D. None of these

Answer 1

The correct option is.

B

$-83.48J$

$\Delta q=-124\text{J}(\because \text{the heat is removed})$

$V_1=377\text{mL}$

$V_2=177\text{mL}$

$\therefore \Delta V=V_2-V_1=177-377=-200\text{mL}=-0.2\text{L}$

$P=1520\text{torr}=\frac{1520}{760}=2\text{atm}$

$P.\Delta V=2\times (-0.2)=-0.4\text{L atm}$

As we know that,

$1\text{L atm}=101.3\text{J}$

$\therefore P.\Delta V=-0.4\text{L atm}=(-0.4\times 101.3)\text{J}=-40.52\text{J}$

As we know that,

$\Delta U=\Delta q-P\Delta V$

$\Delta U=-124-(-40.52)$

$\Rightarrow \Delta U=-124+40.52=-83.48\text{J}$