Question

$2.8\times {10}^{-2}$ kg of nitrogen is expanded isothermally and reversibly at $300K$ from $15.15\times {10}^5{Nm}^{-2}$ when the work done is found to be -17.33 kJ. Find the final pressure.

Answer 1

$W=-2.303nRTlog\left(\frac{P_1}{P_2}\right)$

$-17.33kJ=-2.303\times \frac{W}{M}\times RTlog\frac{P_1}{p_2}$

$-17.33\times {10}^3=\frac{-2.303\times 2.8\times {10}^{-2}\times {10}^3}{28}\times 8.314\times 300\times log\frac{15.15\times {10}^5}{P_2}$

$-17.33\times {10}^3=-2.303\times \frac{2.8\times {10}^{-2}\times {10}^3}{28}\times 8.314\times 300\times (log15.15\times {10}^5-l0g{P}_2)$

$(log15.15\times {10}^5-log{P}_2)=\frac{-17.33\times {10}^3\times 28}{-2.303\times 2.8\times 10\times 8.314\times 300}$

$(6.18-log{P}_2)$=$3.017$

$log{P}_2=6.18-3.017=3.163$

$P_2=antilog\left(3.163\right)$

$P_2=1455.46N/m^2$