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!
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) = ?
So I got 251.5
Is that right?
And thank you so much!
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.
To determine Delta E (the change in internal energy), we need to use the first law of thermodynamics, which states that the change in internal energy is equal to the heat added to the system minus the work done by the system:
Delta E = q - w
Given that q = 773 J (heat absorbed) and w = -PΔV (pressure times change in volume), we can calculate Delta E.
However, the value of w you provided, -(1.977)(2.603), appears to be incorrect. The correct work done by the system can be found by multiplying the change in volume by the external pressure:
w = -PΔV = -(1.977 atm)(6.877 L - 4.274 L)
To perform this calculation, we need to convert the units of atm and L to Joules. The conversion factor is 1 L atm = 101.325 J.
w = -PΔV = -(1.977 atm)(6.877 L - 4.274 L)(101.325 J/L atm)
By evaluating this expression, you will get the correct value for w in Joules.
With the correct values of q and w, you can proceed as follows:
Delta E = 773 J - (correct value of w)
Performing this calculation will give you the final answer for Delta E in Joules.