Pressure, Volume and Temp Change WORK done?

In summary, the conversation discusses the process of heating 9.95 moles of an ideal gas from an initial pressure of 0.922 atm and an initial volume of 679 L to a final pressure of 1.16 atm and a final volume of 1073 L. The initial temperature of the gas is found to be 767 K. The work done on the system is calculated by finding the area under the curve on a P-V diagram, but the incorrect sign was used, resulting in an incorrect answer. The correct work done on the system is 41.5 kJ.
  • #1
sweetpete28
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Pressure, Volume and Temp Change...WORK done?

n = 9.95 moles of an ideal gas are slowly heated from initial pressure P1 = .922 atm and initial volume V1 = 679 L to final pressure Pf = 1.16 atm and final volume Vf = 1073 L. Find T1, the initial temperature of the gas. If the plot of this process on a P-V diagram is a straight line, find W, the work done on the system.

P1 = .922 atm x 1.013 x 10^5 Pa = 93398.6 Pa
V1 = 679 L / 1000 = .679 m^3

PV = nRT
(93398.6)(.679) = (9.95)(8.314)(T)
T = 767 K

For the Work done I would just need to find the area under the curve for the P-V diagram...correct? Well that is what I did and my answer is considered wrong and I cannot understand why...

When V = .679 m^3, P = 93398.6 Pa. And when V = 1.073 m^3, P = 117508 Pa. The area of the rectangle under the curve is (1.073 - .679) x 93398.6 = 36799.0484 J. The area of the triangle under the curve is 1/2bh = (.5)(1.073-.679)(117508 - 93398.6) = 4749.5518 J

So W SHOULD = 36799.0484 + 4749.5518 = 41548.6002 J = 41.5 kJ...but this is considered WRONG...Can someone please advise what I am doing wrong?? Very frustrated here!
 
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  • #2


Check the sign of work. The work done on the system was asked.

ehild
 

Related to Pressure, Volume and Temp Change WORK done?

1. How does pressure affect the work done on a system?

According to the first law of thermodynamics, the work done on a system is equal to the change in internal energy plus the heat added to the system. When the pressure on a system increases, the internal energy of the system also increases, resulting in more work being done on the system.

2. What is the relationship between volume and work done on a gas?

The work done on a gas is directly proportional to the change in volume. This means that as the volume of a gas increases, the work done on the gas also increases. In other words, if the volume of a gas is doubled, the work done on the gas will also double.

3. How does temperature change affect the work done on a system?

The work done on a system is directly proportional to the change in temperature. This means that as the temperature of a system increases, the work done on the system also increases. In other words, if the temperature of a system is doubled, the work done on the system will also double.

4. Can work be done on a system without changing its pressure, volume, or temperature?

Yes, work can be done on a system without changing its pressure, volume, or temperature. This is known as zero work, and it occurs when there is no change in the system's internal energy. An example of this is when a gas expands or contracts in a container with rigid walls.

5. How is work calculated for a gas at constant pressure?

The work done on a gas at constant pressure can be calculated using the formula W = PΔV, where W is the work done, P is the pressure, and ΔV is the change in volume. This formula assumes that the pressure remains constant throughout the process and can be used for both expansion and compression of a gas.

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