Change in internal energy for an isobaric process?

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In an isobaric process involving a monatomic ideal gas, the pressure remains constant while the volume increases from 2.00 x 10^-3 m^3 to 8.00 x 10^-3 m^3 at 4.00 x 10^4 Pa. The work done during this expansion can be calculated using the formula PΔV. To determine the change in internal energy (ΔU), it is essential to know the change in temperature (ΔT) and the number of moles (n) of the gas, as ΔU depends on these variables. The relationship ΔU = Q - W is crucial, where Q is the heat added and W is the work done. Utilizing the ideal gas law can help find the unknowns necessary to calculate ΔU accurately.
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Question:

When a quantity of monatomic ideal gas expands at a constant pressure of 4.00 \times 10^{4} {\rm Pa}, the volume of the gas increases from 2.00 \times 10^{ - 3} {\rm m}^{3} to 8.00 \times 10^{ - 3} {\rm m}^{3}.

A.

What is the change in the internal energy of the gas?

It's isobaric, so the pressure is constant.

I know the work is P\Delta V = (4.00 * 10^4)(6.00 * 10^{-3}).

But, I don't know how to get \Delta U from this.
 
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What variables does the internal energy of the ideal gas depend on? How do these variables change in the aforementioned process?
 
OK, the internal energy depends only on temperature.

For a monatomic gas, \Delta U = \frac{3}{2}nR\Delta T.

I don't know the number of moles or the change in temperature.
 
Progress! Ok, so now you need to know the change in temperature times n R, right? You know the pressure and volume of the gas at two different points in P,V space. Can you use this information to find the unknown? Hint: ideal gas law.
 
This relationship should be helpful as well.

\Delta U = Q - W

Apparently you have the equation for the W right. Now use the ideal gas law and a bit of calorimetry.
 
Thanks a lot!
 
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