Answer:
![P_2=0.398bar=39800Pa](https://tex.z-dn.net/?f=P_2%3D0.398bar%3D39800Pa)
![T_2=118.7K\\](https://tex.z-dn.net/?f=T_2%3D118.7K%5C%5C)
![Q=-3729.9J](https://tex.z-dn.net/?f=Q%3D-3729.9J)
![W=-61753.24J](https://tex.z-dn.net/?f=W%3D-61753.24J)
Δ![U_T=0J](https://tex.z-dn.net/?f=U_T%3D0J)
Δ![H_T=0J](https://tex.z-dn.net/?f=H_T%3D0J)
Explanation:
Hello,
At the first state, the molar volume is:
![v_1=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^6Pa}=2.48x10^{-3}m^3](https://tex.z-dn.net/?f=v_1%3D%5Cfrac%7BRT%7D%7BP_1%7D%20%3D%5Cfrac%7B8.314%5Cfrac%7BPa%2Am%5E3%7D%7BmolK%7D%2A298.15%7D%7B1x10%5E6Pa%7D%3D2.48x10%5E%7B-3%7Dm%5E3)
The volume in both the second and third state:
![v_2=v_3=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^5Pa}=2.48x10^{-2}m^3](https://tex.z-dn.net/?f=v_2%3Dv_3%3D%5Cfrac%7BRT%7D%7BP_1%7D%20%3D%5Cfrac%7B8.314%5Cfrac%7BPa%2Am%5E3%7D%7BmolK%7D%2A298.15%7D%7B1x10%5E5Pa%7D%3D2.48x10%5E%7B-2%7Dm%5E3)
Now, as it is about an adiabatic process, one remembers the following relationships:
![PV^\alpha =K\\TV^{\alpha-1}\\\alpha=\frac{Cp}{Cv}=\frac{7/2R}{5/2R}=1.4](https://tex.z-dn.net/?f=PV%5E%5Calpha%20%3DK%5C%5CTV%5E%7B%5Calpha-1%7D%5C%5C%5Calpha%3D%5Cfrac%7BCp%7D%7BCv%7D%3D%5Cfrac%7B7%2F2R%7D%7B5%2F2R%7D%3D1.4)
- Next, for the aforesaid volumes and the first pressure, one computes the second pressure as:
![P_2=\frac{P_1V_1^\alpha }{V_2^\alpha} =\frac{10bar*(2.48x10^{-3}m^3)^{1.4}}{(2.48x10^{-2}m^3)^{1.4}} =0.398bar=39800Pa](https://tex.z-dn.net/?f=P_2%3D%5Cfrac%7BP_1V_1%5E%5Calpha%20%7D%7BV_2%5E%5Calpha%7D%20%3D%5Cfrac%7B10bar%2A%282.48x10%5E%7B-3%7Dm%5E3%29%5E%7B1.4%7D%7D%7B%282.48x10%5E%7B-2%7Dm%5E3%29%5E%7B1.4%7D%7D%20%3D0.398bar%3D39800Pa)
- And the temperature:
![T_2=\frac{T_1V_1^{\alpha-1}}{V_2^{\alpha-1}} =\frac{298.15K*(2.48x10^{-3}m^3)^{1.4-1}}{(2.48x10^{-2}m^3)^{1.4-1}} =118.7K\\](https://tex.z-dn.net/?f=T_2%3D%5Cfrac%7BT_1V_1%5E%7B%5Calpha-1%7D%7D%7BV_2%5E%7B%5Calpha-1%7D%7D%20%3D%5Cfrac%7B298.15K%2A%282.48x10%5E%7B-3%7Dm%5E3%29%5E%7B1.4-1%7D%7D%7B%282.48x10%5E%7B-2%7Dm%5E3%29%5E%7B1.4-1%7D%7D%20%3D118.7K%5C%5C)
- Q:
It is clear that the heat for the first process is 0 as it is adiabatic, but for the second one, it is computed as:
![Q_2=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J](https://tex.z-dn.net/?f=Q_2%3DnCv%28T_2-T_1%29%3D1mol%2A%5Cfrac%7B5%7D%7B2%7D%288.314%5Cfrac%7BJ%7D%7Bmol%2AK%7D%29%2A%28118.7K-298.15K%29%3D-3729.9J)
Then the total heat:
![Q=Q_1+Q_2=0-3729.9J=-3729.9J](https://tex.z-dn.net/?f=Q%3DQ_1%2BQ_2%3D0-3729.9J%3D-3729.9J)
- The work for the first process is:
![W_1=\frac{P_2V_2-P_1V_1}{1-\alpha }=\frac{39800Pa*2.48x10^{-3}m^3-1x10^6Pa*2.48x10^{-2}m^3}{0.4} \\W_1=-61753.24J](https://tex.z-dn.net/?f=W_1%3D%5Cfrac%7BP_2V_2-P_1V_1%7D%7B1-%5Calpha%20%7D%3D%5Cfrac%7B39800Pa%2A2.48x10%5E%7B-3%7Dm%5E3-1x10%5E6Pa%2A2.48x10%5E%7B-2%7Dm%5E3%7D%7B0.4%7D%20%5C%5CW_1%3D-61753.24J)
It is clear that the second process is isochoric, so the work here is zero, thus, the total work is:
![W=W_1+W_2=-61753.24J+0J=-61753.24J](https://tex.z-dn.net/?f=W%3DW_1%2BW_2%3D-61753.24J%2B0J%3D-61753.24J)
- For the two processes, ΔU becomes the same value since the system returns to the initial temperature, so ΔU total is 0, thus, for each process, one's got:
![U_1=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J\\U_2=nCv(T_3-T_2)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=3729.9J\\](https://tex.z-dn.net/?f=U_1%3DnCv%28T_2-T_1%29%3D1mol%2A%5Cfrac%7B5%7D%7B2%7D%288.314%5Cfrac%7BJ%7D%7Bmol%2AK%7D%29%2A%28118.7K-298.15K%29%3D-3729.9J%5C%5CU_2%3DnCv%28T_3-T_2%29%3D1mol%2A%5Cfrac%7B5%7D%7B2%7D%288.314%5Cfrac%7BJ%7D%7Bmol%2AK%7D%29%2A%28298.15K-118.7K%29%3D3729.9J%5C%5C)
- Finally, the total enthapy is also 0 due to same aforesaid reason, thus, each enthalpy is:
![H_1=nCp(T_2-T_1)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-5221.86J\\H_2=nCv(T_3-T_2)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=5221.86J\\](https://tex.z-dn.net/?f=H_1%3DnCp%28T_2-T_1%29%3D1mol%2A%5Cfrac%7B7%7D%7B2%7D%288.314%5Cfrac%7BJ%7D%7Bmol%2AK%7D%29%2A%28118.7K-298.15K%29%3D-5221.86J%5C%5CH_2%3DnCv%28T_3-T_2%29%3D1mol%2A%5Cfrac%7B7%7D%7B2%7D%288.314%5Cfrac%7BJ%7D%7Bmol%2AK%7D%29%2A%28298.15K-118.7K%29%3D5221.86J%5C%5C)
Best regards.