(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

0.21 x 10^{-3}kmol of ideal gas occupy a volume of 5.0 x 10^{-3}m^{3}initiallyat a pressure of 1.2 x 10^{5}Pa and temperatureT.

The volume of the gas is first decreased to 2.0 x10^{-3}m^{3}at a constant pressure of 1.2 x 10^{5}Pa, and then the pressure is increased to 3.0 x10^{5}Pa at a constant volume of 2.0 x10^{-3}m^{3}.

For this process, calculate:

A) The Work done on the gas

B) The Heat absorbed by the gas

2. Relevant equations

ΔU = Q - W

(where ΔU is change in internal energy, Q is Heat and W is work donebythe gas)

U = n N_{A}([itex]\frac{1}{2} m \overline{v^2}[/itex]) = [itex]\frac{3}{2} R T[/itex]

where U is internal energy

W = p ( V_{2}- V_{1})(for isobaric process)

W = 0(for isovolumetric process)

3. 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 (V_{2}- V_{1}) = - 1.2 x 10^{5}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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# Homework Help: Heat Absorbed during a process on an Ideal Gas

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