OUNT OF HEAT ABSORBED AND WORK DONE IN A PV DIAGRAM?

AI Thread Summary
The discussion focuses on calculating the work done by a gas and the heat absorbed during a process represented on a PV diagram. The initial state of the gas is at 1.00 atm and 14.0 L, transitioning to a final state of 2.00 atm and 78.0 L. Participants clarify that the area under the PV curve is not a triangle, emphasizing the need to calculate the work done correctly as the area down to P=0. Additionally, there is a reminder to convert volume units appropriately and to consider the relationship between work, heat flow, and internal energy changes. The conversation highlights the importance of understanding these thermodynamic principles for accurate calculations.
paradiselost5
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Homework Statement



One mole of a dilute gas initially has a pressure equal to 1.00 atm, and a volume equal to 14.0 L. As the gas is slowly heated, the plot of its state on a PV diagram moves in a straight line to the final state. The gas now has a pressure equal to 2.00 atm, and a volume equal to 78.0 L.

(a) Find the work done by the gas.


(b) Find the heat absorbed by the gas.

Homework Equations





The Attempt at a Solution


I took the area under the PV curve which is a triangle... 64/2*101.325
 
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paradiselost5 said:


The Attempt at a Solution


I took the area under the PV curve which is a triangle... 64/2*101.325

Watch out: It is not a triangle. The work done by the gas is the area under the PV curve down to P=0.

ehild
 
In addition to what ehild points out, watch the units for volume. If you are using MKS units you have to convert litres to _________.

What is the relationship between work done, heat flow and change in internal energy? To find the heat flow, what other quantity do you have to find? (hint: what is the change in temperature?).

AM
 
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