Answer to Question #214356 in Molecular Physics | Thermodynamics for Wavie

Question

Calculate the internal energy and enthalpy of 1 kg of air occupying 0.05m3 at 20 bar. If the internal energy is increased by 120 kJ/kg as air is compressed to 50 bar, calculate the new volume occupied by 1 kg of air. Assume air is a perfect gas

Answer 1

Solution.

$m=1kg;$

$V_1=0.05m^3;$

$p_1=2\sdot10^6Pa;$

$\Delta U=120\sdot10^3J;$

$p_2=5\sdot10^6Pa;$

$M=28.96\sdot10^{-3}kg/mol;$

$T_1=293K;$

$U=\dfrac{m}{M}C_VT;$

$C_V=\dfrac{i}{2}R=\dfrac{5}{2}\sdot8.31=20.8J/molK;$

$U=\dfrac{1}{28.96\sdot10^{-3}}\sdot20.8\sdot293=210442J;$

$H=U+pV;$

$H=210442+2\sdot10^6\sdot0.05=310442J;$

$\Delta U=\dfrac{m}{M}C_V(T_2-T_1)\implies T_2=\dfrac{\Delta U M}{mC_V}+T_1;$

$T_2=\dfrac{120\sdot10^3\sdot28.96\sdot10^{-3}}{1\sdot20.8}+293=460K;$

$\dfrac{P_1\sdot V_1}{T_1}=\dfrac{P_2\sdot V_2}{T_2}\implies V_2=\dfrac{P_1V_1T_2}{T_1P_2};$

$V_2=\dfrac{2\sdot10^6\sdot0.05\sdot460}{293\sdot5\sdot10^6}=0.031m^3;$

Answer: $U=210442J;$

$H=310442J;$

$V_2=0.031m^3.$