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Three moles of an ideal gas are taken around the eyele acb shown in Fig.For this gas Co-29.2J/mol-K.Process ac is at constant pressure, process ba is at constant volume, and process cb is adiabatic. The temperatures of the gas in states a c, and b are T-300KTb600K and T-492KCalculate the total work Ffor the cycle.(R=8.31J/molK)

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okpalawalter8

For three moles of an ideal gas are taken around the eye, the total work for the cycle is mathematically given as

Wt=1945.475J

What is the total work F for the cycle.(R=8.31J/molK)?

Generally, the equation for work  is mathematically given as

Wt=wac+Wc+Wba

Therefore

Wac=Pa(Vc-Va)=nR(Tc-Ta)

Wac=3(8.314*192)

Wac=4788.864J

Wcb=P1v1-p2v2/v-1

Wcb-3*8.34*108/0.4

Wcb=-6734.34J

Wab=0

In conclusion,

Wt=wac+Wc+Wba

Wt=4788.864J+0+(-6734.34J)

Wt=+1945.475J

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