Question

A gas is allowed to expand at constant temperature from 4.274L to 6.877L against an external pressure of 1.977 atm. If at the same time the gas absorbs 773 J of heat from the surroundings, what is Delta E for this process? (Note: Pay attention to units; 1 L atm = 101.325 J)

I got
q=773J
w=-(1.977)(2.603)
Delta E = 768

Is this right?
I am looking more for the process than an actual answer here!! Thank you!

Answer 1

No. Note the problem specifically suggests you pay attention to the units and you didn't do that. If you use 1.977(2.603) that gives you units for work in L*atma and you want J here. You convert L*atm to J by L*atm x 101.325 = J. Your sign of - is correct. The sign for 773 is correct at + also.
Therefore dE = 773 -(1.977)(2.603)(101.325) = ?

Answer 2

So I got 251.5
Is that right?
And thank you so much!

Answer 3

I obtained 251.568. I don't know how many significant figures you have on that 773 but everything else you have is 4 places. If you can have 4 s.f. I would round that to 251.6 J.