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=m\,s\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 $