## In This post you will get answer of Q.

$ 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

Solution:

## 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}} $

$ = 273left( frac{5}{1}right)^{frac{14-1}{1.4}} $

$ = 273times (5)^{2/7} $

$ = 273times 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 $