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Heat Absorbed during a process on an Ideal Gas

  • Thread starter adammw111
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Homework Statement



0.21 x 10-3 kmol of ideal gas occupy a volume of 5.0 x 10-3 m3 initially at a pressure of 1.2 x 105 Pa and temperature T.

The volume of the gas is first decreased to 2.0 x10-3 m3 at a constant pressure of 1.2 x 105 Pa, and then the pressure is increased to 3.0 x105 Pa at a constant volume of 2.0 x10-3 m3.

For this process, calculate:

A) The Work done on the gas

B) The Heat absorbed by the gas

Homework Equations



ΔU = Q - W
(where ΔU is change in internal energy, Q is Heat and W is work done by the gas)

U = n NA ([itex]\frac{1}{2} m \overline{v^2}[/itex]) = [itex]\frac{3}{2} R T[/itex]
where U is internal energy

W = p ( V2 - V1 ) (for isobaric process)
W = 0 (for isovolumetric process)

The Attempt at a Solution



A rough P-V diagram would look like:

[PLAIN]http://img695.imageshack.us/img695/5465/pvdiagram.png [Broken]

A) Work done on the gas.

Work done is area under P-V curve.

W = p (V2 - V1) = - 1.2 x 105 x (2.0 x 10-3 - 5.0 x 10-3) = + 360 J

B) Heat absorbed by the gas.

ΔU = Q - W, therefore:
Q = ΔU + 360

However I don't know how to find ΔU and seem to be stuck.
 
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Answers and Replies

  • #2
ehild
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You can calculate the internal energy from the temperature. Use the ideal gas law. How does the temperature change between the initial and final states?

ehild
 
  • #3
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Using [itex]\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}[/itex] (ideal gas law):
[itex]\frac{1.2 \times 10^5 \times 5.0 \times 10^-3}{T_{initial}} = \frac{3.0 \times 10^5 \times 2.0 \times 10^-3}{T_{final}}[/itex]
[itex]600 T_{initial} = 600 T_{final}[/itex]
[itex]T_{initial} = T_{final}[/itex]

Therefore as an isothermic process (constant temperature), ΔU = 0, Q=W.
So Q = 360 J.

Is this correct? The answers I have for this sample exam give -360 J instead.
 
  • #4
ehild
Homework Helper
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W is the work of the gas. As the volume decreases in the first step, the work of the gas is negative, W=-360 J. At the same time, Q is the adsorbed heat. ΔU=Q-W(gas)=Q-(-360J)=0 --->Q=-360 J. Some external force does work on the gas, and the gas transfers heat to the surroundings.


ehild
 
  • #5
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I get it now...
I was getting confused because the question asked work done on the gas when my formulas are for work done by the gas, and so I used the wrong number/sign in the second part. Your explanation makes sense.

Thanks, ehild.
 
  • #6
ehild
Homework Helper
15,427
1,825
I get it now...
I was getting confused because the question asked work done on the gas when my formulas are for work done by the gas, and so I used the wrong number/sign in the second part. Your explanation makes sense.

Thanks, ehild.
You are welcome.

The First Law of Thermodynamics states that the internal energy increases by the amount of heat added to and by the work done on the gas: ΔU=Q+W. If the work is done very slowly, so the gas is very nearly in equilibrium with the external force, W=-PdV=-Wgas.

ehild
 

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