Question

$ 50 \,g $ of oxygen at $ NTP $ is compressed adiabatically to a pressure of $ 5 $ atmosphere. The work done on the gas, if $ \gamma = 1.4 $ and $ R = 8.31\, J\, mol^{-1}\, K^{-1} $ is

• A. $ -5173\,J $
• B. $ 1131\,J $
• C. $ -1364\,J $
• D. $ 5673\,J $

Answer 1

Answer: (A)

Her, $ T_1 = 273 \,K, P_1 = 1 $ atm, $ P_2 = 5 $ atm
no. of moles of oxyzen $ = \frac{50}{32} $
$ \Rightarrow T_{2} = T_{1} \left(\frac{P_{1}}{P_{2}}\right)^{\frac{\gamma -1}{\gamma}} $
$ = 273\left( \frac{5}{1}\right)^{\frac{14-1}{1.4}} $
$ = 273\times (5)^{2/7} $
$ = 273\times 1.584 $
$ = 432.37 $
$ W = \frac{nR}{\gamma-1}[T_{1} - T_{2} ] $
$ = \frac{50}{32}\times \frac{8.31}{1.4 -1} (273-432.37) $
$ = -5173\,J $