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## Homework Statement

0.21 x 10

^{-3}kmol of ideal gas occupy a volume of 5.0 x 10

^{-3}m

^{3}

**initially**at a pressure of 1.2 x 10

^{5}Pa and temperature

*T*.

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

## 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 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)*

## 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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