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### Topic: First law of thermodynamics  (Read 399 times)

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#### eitan123

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##### First law of thermodynamics
« on: October 02, 2020, 05:59:17 AM »
2 moles of an ideal gas are in a closed system with a temperature of 25°C.

We cool the gas under constant pressure and then heat the gas under constant volume.

During the cooling process the gas has lost 1000J of heat to the environment.

It is known that the molar heat capacity of the gas is 25J/(k°*mole)

What is the absolute value of the total work done by the system in the cooling and heating combined?

What I tried was to apply the first law of thermodynamics.

It is known that Q=-1000J.

Then I found ΔE. To do that I calculated the temperature difference:

-1000=2*25*ΔT, therefore ΔT=-20°C

Then because the gas is ideal we may say the ΔE=1.5nRΔT, that gives ΔE=1.5*2*8.314*-20=-498.84J.

The using the first law we get that W=501.16J.

Because the heating is done under constant volume there shouldn't be any work done there.

As you can see my answer does not match the official answer, where is my mistake?

#### Enthalpy

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##### Re: First law of thermodynamics
« Reply #1 on: October 02, 2020, 05:11:25 PM »
25J/mol/K is not compatible with 3/2RT.

3/2RT is the internal energy of a monoatomic ideal gas. In many texts, ideal gases can be polyatomic - and this makes even sense in real life.

You should also double-check if 25J/mol/K is at constant volume or pressure.

As a side note, the symbol for kelvin is K.