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## Homework Statement

An ideal gas(γ=1.4) was expanded under constant pressure. The work done was 80 Joule.Calculate the heat absorbed and increase in internal energy of the system.

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- Thread starter dk_ch
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An ideal gas(γ=1.4) was expanded under constant pressure. The work done was 80 Joule.Calculate the heat absorbed and increase in internal energy of the system.

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[itex]w= P \int dV[/itex]

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[itex]w= P \int dV[/itex]

what is the solution?

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Start by thinking about this: is the change in energy path dependent or path independent?

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Start by thinking about this: is the change in energy path dependent or path independent?

W = Q-ΔU is known from from first law of thermodynamics and w is given here . Now how shall i proceed using γ=1.4 ? how to apply your equation ?

please help.

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- #6

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Is this calculus based physics or algebra based?

See this is why you need to show some work, I have no idea at what level you're doing this. If the integral is what's confusing you then simply change it to: [itex]w = P \Delta V[/itex].

You want to find [itex]Q_{in}[/itex] for a constant pressure problem.

[itex]\Delta E = W + Q[/itex]

We know that w= -80J. Because the system did work on the piston it has to be negative - energy left the system.

[itex]\Delta E + 80 = Q[/itex]

[itex]\gamma = 1.4[/itex] Therefore you know it's a diatomic gas.

And I actually get stuck there too.. Hmm. You aren't given either moles of gas or change in temperature? With your gamma you can find Cp and Cv but you don't seem to have enough information.

[itex]Q = nC_{p} \Delta T[/itex]

and [itex]\Delta E = nC_{v} \Delta T[/itex]

Edit: It's ambiguous if they mean that the work done was by the system or on the system. Is this the whole original question?

See this is why you need to show some work, I have no idea at what level you're doing this. If the integral is what's confusing you then simply change it to: [itex]w = P \Delta V[/itex].

You want to find [itex]Q_{in}[/itex] for a constant pressure problem.

[itex]\Delta E = W + Q[/itex]

We know that w= -80J. Because the system did work on the piston it has to be negative - energy left the system.

[itex]\Delta E + 80 = Q[/itex]

[itex]\gamma = 1.4[/itex] Therefore you know it's a diatomic gas.

And I actually get stuck there too.. Hmm. You aren't given either moles of gas or change in temperature? With your gamma you can find Cp and Cv but you don't seem to have enough information.

[itex]Q = nC_{p} \Delta T[/itex]

and [itex]\Delta E = nC_{v} \Delta T[/itex]

Edit: It's ambiguous if they mean that the work done was by the system or on the system. Is this the whole original question?

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- #7

CAF123

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And I actually get stuck there too.. Hmm. You aren't given either moles of gas or change in temperature? With your gamma you can find Cp and Cv but you don't seem to have enough information.

[itex]Q = nC_{p} \Delta T[/itex]

and [itex]\Delta E = nC_{v} \Delta T[/itex]

Divide these two equations and you will get Q/ΔE = c

- #8

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Ah! Genius. I would never have thought of that :).Divide these two equations and you will get Q/ΔE = c_{p}/c_{v}= γ = 1.4. Along with the equation from conservation of energy, you now have two equations with two unknowns.

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Divide these two equations and you will get Q/ΔE = c_{p}/c_{v}= γ = 1.4. Along with the equation from conservation of energy, you now have two equations with two unknowns.

well done. Thanks to both of you.

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