- #1

Tehy

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5 mol of gas is in temperature of 326K and the same time the pressure is expanded from 1atm to 3atm.

**How much work is done by the gas?**

Detailed instruction would be good, thanks :)

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- Thread starter Tehy
- Start date

- #1

Tehy

- 7

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5 mol of gas is in temperature of 326K and the same time the pressure is expanded from 1atm to 3atm.

Detailed instruction would be good, thanks :)

- #2

sicjeff

- 46

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W/mole = dPV = Pdv + Vdp.

it should be pretty obvious from here.

it should be pretty obvious from here.

- #3

Tehy

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W/mole = dPV = Pdv + Vdp.

it should be pretty obvious from here.

Yes, it probably should, but I'm very dumb with physics :P I already tried to find some help from Google but I didn't have any luck :/

- #4

sicjeff

- 46

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Let's assume an ideal gas.

we know the standard molar volume right? (22.414 L/mole). so v=nRT/P.

We are assuming that the size of our container is constant. (isochoric condition)

- #5

Tehy

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Let's assume an ideal gas.

we know the standard molar volume right? (22.414 L/mole). so v=nRT/P.

We are assuming that the size of our container is constant. (isochoric condition)

Oh, yes it's ideal gas. I forget to mention that :)

Do I need to calculate v1= nRT/1atm and then v2= nRT/3atm?

- #6

Andrew Mason

Science Advisor

Homework Helper

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The question does not have sufficient information. You have to know how the pressure is changed and whether heat is being added or released.

5 mol of gas is in temperature of 326K and the same time the pressure is expanded from 1atm to 3atm.How much work is done by the gas?

Detailed instruction would be good, thanks :)

If the external pressure is increased, the volume must decrease, in which case the work done by the gas is negative. If pressure is increased because heat is added, the volume can be held constant, in which case, no work is done. Or it could increase, in which case the work done by the gas is positive.

Work = Pdv not [itex]\Delta (PV)[/itex]

If you assume that the external pressure is gradually changed from 1 atm to 3 atm, and the temperature is kept constant, the work done is:

[tex]W = \int_{P_i}^{P_f} PdV = \int_{P_i}^{P_f} nRTdV/V = nRT\ln(\frac{V_f}{V_i}) = -nRT\ln(\frac{V_i}{V_f})[/tex]

If it is adiabatic (temperature will increase) it is more complicated.

AM

- #7

Tehy

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Thanks Andrew! That works great! :)

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