Question

You have gas in a container with a movable piston. The walls of the container are thin enough so that its temperature stays the same as the temperature of the surrounding medium. You have baths of water of different temperatures, different objects that you can place on top of the piston, etc. You make the gas undergo an isothermal process so that the pressure inside increases by 10%, then undergo an isobaric process so that the new volume decreases by 20%, and finally undergo an isochoric process so that the temperature increases by 15%. What is the new pressure of the Gas? What is the new volume of the gas? What is the new temperature of the gas?

Answer 1

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:

$PV=nRT$

• For Isothermal process:

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$(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:

$V_{2}=\frac{V_{1}}{1.1}$ (2)

• For Isobaric process:

$\frac{P_{2}V_{2}}{T_{2}}=\frac{P_{3}V_{3}}{T_{3}}$ (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:

$T_{3}=0.8T_{2}$ (4)

• For Isochoric process:

$\frac{P_{3}V_{3}}{T_{3}}=\frac{P_{4}V_{4}}{T_{4}}$ (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:

$P_{4}=1.15P_{3}$ (6)

So now because P4=1.15*P3, P2=P3 and P2=1.1*P1:

$P_{4}=1.15*1.1P_{1}=1.265P1$

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:

$V_{4}=\frac{0.8}{1.1}V_{1}=0.72V1$

so the final volume decreases 27% with respect to initial volume.

T4=1,15*T3, T3=0.8*T2 and T1=T2:

$T_{4}=1.15*0.8T_{1}=0.92T1$

The new temperature decreases 8% with respect to initial volume.