Question

One mole of an ideal gas for which $C_V=\frac{3}{2}R$ is heated reversibly at a constant pressure of 1 atm from ${25}^{o}C$ to ${100}^{o}C$ , the $\Delta H$ is

• A. $3.775cal$
• B. $37.256cal$
• C. $372.56cal$
• D. $3725.6cal$

Answer 1

Answer: (C)

Key Idea: We know that $C_P-C_V=R$ or $C_P=C_V+R$ $\therefore$ $C_V=\frac{3}{2}R$ $\because$ $C_P=\frac{3}{2}R+R=\frac{5}{2}R$ At constant pressure heat given to one mole gas $\Delta H=ms\Delta T$ $\therefore$ $\Delta H=q_p=1\times \frac{5}{2}R\times (373-298)$ $=1\times \frac{5}{2}\times 1.987\times 75$ $=372.56cal$