Question

A gas is compressed at a constant pressure of $50{\text{N/m}}^2$ from a volume of $10{\text{m}}^3$ to a volume of $4{\text{m}}^3$. Heat of $100\text{J}$ is then added to the gas by heating. Its internal energy is

• A. Decreased by $200\text{J}$
• B. Increased by $200\text{J}$
• C. Decreased by $400\text{J}$
• D. Increased by $400\text{J}$

Answer 1

The correct option is D Increased by $400\text{J}$

Given,
Energy added to the system, $\Delta Q=+100J$
Pressure applied $\left(P\right)=50{\text{N/m}}^2$
Change in volume, $\Delta V=V_2-V_1=4-10=-6m^3$
From first law of the thermodynamics
$\Delta Q=\Delta U+\Delta W$
we can write that,
$\Delta Q=\Delta U+P\Delta V$
$\Rightarrow +100=\Delta U+50\times (-6)$
$\Rightarrow \Delta U=400\text{J}$
Hence , option 'D' is the correct answer.