Answer:
New pressure of the gas increases by 26.5% with respect to initial pressure, new volume decreases 27% with respect to initial volume and new temperature decreases 8% with respect to initial volume.
Explanation:
If we assume the gas is a perfect gas we can use the perfect gas equation:
![PV=nRT](https://tex.z-dn.net/?f=PV%3DnRT%20)
(1)
Where subscripts 1 shows before the isothermal process and 2 after it, because isothermal means constant temperature T1=T2, and pressure increases by 10% means P2=1,1*P1, using these facts on (1) we have:
(2)
(3)
Where subscripts 2 shows before the isobaric process and 3 after it, because isobaric means constant pressure P2=P3, and volume decreases by 20% means V3=0.8*V2, using these facts on (3) we have:
(4)
(5)
Where subscripts 3 shows before the isochoric process and 4 after it, because isochoric means constant volume V3=V4, and temperature increases by 15% means T4=1.15*T3, using these facts on (5) we have:
(6)
So now because P4=1.15*P3, P2=P3 and P2=1.1*P1:
![P_{4}=1.15*1.1P_{1}=1.265P1](https://tex.z-dn.net/?f=P_%7B4%7D%3D1.15%2A1.1P_%7B1%7D%3D1.265P1)
This is, the new pressure of the gas increases by 26.5% with respect to initial pressure.
Similarly, we have V3=V4, V3=0.8*V2 and V1=1,1*V2:
![V_{4}=\frac{0.8}{1.1}V_{1}=0.72V1](https://tex.z-dn.net/?f=V_%7B4%7D%3D%5Cfrac%7B0.8%7D%7B1.1%7DV_%7B1%7D%3D0.72V1)
so the final volume decreases 27% with respect to initial volume.
T4=1,15*T3, T3=0.8*T2 and T1=T2:
![T_{4}=1.15*0.8T_{1}=0.92T1](https://tex.z-dn.net/?f=T_%7B4%7D%3D1.15%2A0.8T_%7B1%7D%3D0.92T1)
The new temperature decreases 8% with respect to initial volume.