Calculating The Energy Changes When Heating An Ideal Gas

[Vivien Ho 3L]

I do not really understand the concept in Example 7.6 in the textbook:

Calculate the final temperature and the change in internal energy when 500.J of energy is transferred as heat to 0.900 mol O2(g) at 298K and 1.00 atm at
(a) constant volume
(b) constant pressure. Treat the gas as ideal.

Why do the calculations for constant pressure have to be broken down into two steps?

Can someone please explain the whole concept behind this. Thank you!

[Jennifer Aguayo 1I]

Hi Vivien,
In this problem we are not given the final temperature and we need to know the final temperature in order to find the change in temperature and then solve for delta U.
That is why the first step is to find delta T so then we can plug it in to the delta U=q equation.

I hope this helps.

[Vivien Ho 3L]

Hi Jennifer,

That was not the question I was asking, but thanks anyway!

What I meant was why must the expansion process be broken down into two steps? (ie. heating at constant volume to the final temperature and allowing the gas to expand isothermally?)

Why does the isothermal expansion in this question have a deltaU=0?

[Jenna Kovsky 1I]

If you look at page 27 in the course reader, you see that we're given the equation deltaU=3/2*nRdeltaT for an ideal gas. It follows that if the expansion is happening isothermally, then since deltaT is 0, deltaU is as well. So deltaU is zero for the isothermal expansion of any ideal gas.
If you're confused about the equation, then they derive it in section 7.7 of the textbook, so you could look back at that for clarification.
I hope that this at least partially answers your question!

[Chem_Mod]

At constant volume: deltaU = 5/2nR*deltaT = q. Since the problem gives q, it is easy to find deltaT and deltaU.
At constant pressure: deltaH = 7/2nR*deltaT = q. From this you find deltaT and plug right back into deltaU = 5/2nR*deltaT.

We have used the fact that for a diatomic ideal gas, Cv=5/2R and Cp=7/2R