Question

$4.4\times {10}^{-2}$ kg of $C{O}_2$ are compressed isothermally and reversibly at 293 K from 50kPa when the work obtained is 1.245 kJ. Find the final pressure. (R=8.314J/K).

Answer 1

Amount of $C{O}_2=4.4\times {10}^{-2}kg$

Number of mole of $C{O}_2$

${}^{n}C{o}_2=\frac{4.4\times {10}^{-2}\times {10}^3}{44}$

$=1mole$

work done in isothermal pracess

$W_{iso}=2.303RT{\log }_{10}\frac{V_2}{V_1}=2.303RT{\log }_{10}\frac{P_1}{P_2}$

$1.245\times {10}^3=2.303\times 8.3\times 293{\log }_{10}\frac{P_1}{P_2}$

${\log }_{10}dfrac{P}_1{P}_2=\frac{1295}{2.303\times 8.3\times 293}$

${\log }_{10}\frac{P_1}{P_2}=0.22229$

$\frac{P_1}{P_2}=1.668$

$\frac{50kPa}{1.668}=P_2\Rightarrow P_2=29.77=30kPa$