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

Question #214356

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



1
Expert's answer
2021-07-07T08:39:32-0400

Solution.

m=1kg;m=1kg;

V1=0.05m3;V_1=0.05m^3;

p1=2106Pa;p_1=2\sdot10^6Pa;

ΔU=120103J;\Delta U=120\sdot10^3J;

p2=5106Pa;p_2=5\sdot10^6Pa;

M=28.96103kg/mol;M=28.96\sdot10^{-3}kg/mol;

T1=293K;T_1=293K;

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

CV=i2R=528.31=20.8J/molK;C_V=\dfrac{i}{2}R=\dfrac{5}{2}\sdot8.31=20.8J/molK;

U=128.9610320.8293=210442J;U=\dfrac{1}{28.96\sdot10^{-3}}\sdot20.8\sdot293=210442J;

H=U+pV;H=U+pV;

H=210442+21060.05=310442J;H=210442+2\sdot10^6\sdot0.05=310442J;

ΔU=mMCV(T2T1)    T2=ΔUMmCV+T1;\Delta U=\dfrac{m}{M}C_V(T_2-T_1)\implies T_2=\dfrac{\Delta U M}{mC_V}+T_1;

T2=12010328.96103120.8+293=460K;T_2=\dfrac{120\sdot10^3\sdot28.96\sdot10^{-3}}{1\sdot20.8}+293=460K;

P1V1T1=P2V2T2    V2=P1V1T2T1P2;\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};

V2=21060.054602935106=0.031m3;V_2=\dfrac{2\sdot10^6\sdot0.05\sdot460}{293\sdot5\sdot10^6}=0.031m^3;

Answer: U=210442J;U=210442J;

H=310442J;H=310442J;

V2=0.031m3.V_2=0.031m^3.


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