Question

8 grams of oxygen at STP is compressed adiabatically to a pressure of 10 atm .How much work id done on the gas

Answer 1

For adiabatic process

$A=\frac{m}{M} \frac{RT_1}{\gamma-1}(1-(\frac{V_1}{V_2})^{\gamma-1})=\frac{m}{M} \frac{RT_1}{\gamma-1}(1-((\frac{p_2}{p_1})^{\frac{1}{\gamma}})^{\gamma-1})$

$A=\frac{m}{M} \frac{RT_1}{\gamma-1}(1-((\frac{p_2}{p_1})^{\frac{\gamma-1}{\gamma}}))$

$\gamma=\frac{i+2}{i}=\frac{5+2}{5}=1.4$

So, we have

$A=\frac{m}{M} \frac{RT_1}{\gamma-1}(1-((\frac{p_2}{p_1})^{\frac{\gamma-1}{\gamma}}))=$

$=\frac{0.008}{0.032} \frac{8.31\cdot 273}{1.4-1}(1-((\frac{10\cdot 101325}{101325})^{\frac{1.4-1}{1.4}}))\approx-1320J$