Question

1 kg of nitrogen (molar mass 28 kg/kmol) is compressed reversibly and isothermally from 1.01 bar , 20℃ to 4.2 bar. Calculate the work done and the heat flow during the process. Assume nitrogen to be a perfect gas.

Answer 1

Solution.

$m=1kg;$

$M=28\sdot10^{-3} kg/mol;$

$P_1=1.01bar=101000Pa;$

$T=293K;$

$P_2=4.2bar=420000Pa;$

$A=\dfrac{m}{M}RTln\dfrac{V_2}{V_1};$

$P_1V_1=\dfrac{m}{M}RT\implies V_1=\dfrac{m}{M}\dfrac{RT}{P_1};$

$P_2V_2=\dfrac{m}{M}RT\implies V_2=\dfrac{m}{M}\dfrac{RT}{P_2};$

$\dfrac{V_2}{V_1}=\dfrac{P_1}{P_2};$

$A=\dfrac{1}{28\sdot10^{-3}}\sdot8.31\sdot293\sdot ln\dfrac{101000}{420000}=-124099J=-124kJ;$

Answer: $A=-124kJ;$

theat is rejected.