an ideal gas of monotonic molecules expands from its initial state A to a state B through an isobaric process and then further expands to a volume C . Find the work done by the gas ,increase in internal energy, and the energy transferred by heat to the gas over the whole process
AAB=p1ΔV=4⋅1.49=5.96 J,A_{AB}=p_1\Delta V=4\cdot 1.49=5.96~J,AAB=p1ΔV=4⋅1.49=5.96 J,
ABC=p2V2γ−1(1−(V2V3)γ−1)=4⋅3.492/3⋅(1−(3.498)2/3)=8.90 J,A_{BC}=\frac{p_2V_2}{\gamma -1}(1-(\frac{V_2}{V_3})^{\gamma -1})=\frac{4\cdot 3.49}{2/3}\cdot(1-(\frac{3.49}{8})^{2/3})=8.90~J,ABC=γ−1p2V2(1−(V3V2)γ−1)=2/34⋅3.49⋅(1−(83.49)2/3)=8.90 J,
A=5.96+8.90=14.86 J,A=5.96+8.90=14.86~J,A=5.96+8.90=14.86 J,
ΔUAB=32p1ΔV=8.94 J,\Delta U_{AB}=\frac 32 p_1 \Delta V=8.94~J,ΔUAB=23p1ΔV=8.94 J,
ΔUBC=−ABC=−8.90 J,\Delta U_{BC}=-A_{BC}=-8.90~J,ΔUBC=−ABC=−8.90 J,
ΔU=8.94−8.90=0.04 J,\Delta U=8.94-8.90=0.04~J,ΔU=8.94−8.90=0.04 J,
QAB=AAB+ΔUAB=5.96+8.94=14.90 J,Q_{AB}=A_{AB}+\Delta U_{AB}=5.96+8.94=14.90~J,QAB=AAB+ΔUAB=5.96+8.94=14.90 J,
QBC=0 J,Q_{BC}=0~J,QBC=0 J,
Q=14.90+0=14.90 J.Q=14.90+0=14.90~J.Q=14.90+0=14.90 J.
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
I really appreciate your help
Comments
I really appreciate your help