Answer:
Work transfer W =0.475 kJ
Heat transfer Q=3.68 KJ
Explanation:
At initial condition:
[tex]P_1=2 bar,T_1=200 K,V_1=1 L[/tex]
At final condition:
[tex]P_2=7.5 bar,V_2=2 L[/tex]
Given that pressure -volume follows linear relationship and we know that work for closed system if given as
[tex]w=\int Pdv[/tex]
It means that are of P-v diagram will us work in that process.So we will find the are of P-v diagram.
Now from P-v diagram
Area =[tex]\dfrac{1}{2}(P_1+P_2)(V_2-V_1)[/tex]
=[tex]\dfrac{1}{2}(2+7.5)\times 10^{2}(2-1)\times 10^{-3}[/tex]
=0.475 KJ
So Work transfer W =0.475 kJ
If we assume that air is ideal gas PV=mRT
[tex]\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}[/tex]
[tex]\dfrac{2\times 1}{200}=\dfrac{7.5\times 2}{T_2}[/tex]
[tex]T_2=1500 k[/tex]
From first law of thermodynamics
[tex]Q=U_2-U_1+W[/tex] (For ideal gas U=[tex]mC_v(T_2-T_1)[/tex])
For mass [tex]P_1V_1=mRT_1[/tex]
[tex]Q=m\times 0.71(T_2-T_1)+W[/tex]
[tex]Q=3.48\times 10^{-3}\times 0.71(1500-200)+0.475[/tex]
So heat transfer Q=3.68 KJ
