Question

5 moles of an ideal gas at $100,K$ are allowed to undergo reversible adiabatic compression till its temperature becomes $200,K$. If $C_v=30,J,K^{-1}mol^{-1}$, calculate $\Delta U$ and $\Delta (pV)$ for this process. $(R=8.0,J,K^{-1}mol^{-1})$

• A. $\Delta U=10,kJ;\ \Delta(pV)=2,kJ$
• B. $\Delta U=12,kJ;\ \Delta(pV)=4,kJ$
• C. $\Delta U=14,kJ;\ \Delta(pV)=4,kJ$
• D. $\Delta U=15,kJ;\ \Delta(pV)=4,kJ$
• E. $\Delta U=3.0,kJ;\ \Delta(pV)=2,kJ$

Hint:

Chemistry Tip: For ideal gas, internal energy depends only on temperature, not path.

Answer 1

The Correct Option is D

Concept:
For an ideal gas: 1. Change in internal energy: $$\Delta U=nC_v\Delta T$$ 2. Also: $$pV=nRT$$ So change in $pV$ is: $$\Delta(pV)=nR\Delta T$$
Step 1: Find temperature change.
Initial temperature: $$T_1=100K$$ Final temperature: $$T_2=200K$$ Thus: $$\Delta T = T_2-T_1=100K$$
Step 2: Calculate internal energy change.
Given: $$n=5,\quad C_v=30,J,mol^{-1}K^{-1}$$ $$\Delta U=5\times30\times100$$ $$=15000J=15kJ$$
Step 3: Calculate change in $pV$.
$$\Delta(pV)=nR\Delta T$$ $$=5\times8\times100$$ $$=4000J=4kJ$$
Step 4: Final answer.
$$\boxed{\Delta U=15kJ,\quad \Delta(pV)=4kJ}$$ Hence correct option is (D). :contentReference[oaicite:0]{index=0}