A gas is allowed to expand at constant temperature from a volume of 2.00 L to 11.20 L against an external pressure of 1.500 atm. If the gas loses 256 J of heat to the surroundings, what are the values of q, w, and ΔU? Pay careful attention to units.
-----My attempt:
q = heat
so if the gas losses 256J of heat then that makes
q=-256J
as for w
w = -p (delta V)
w = -( 1.5atm) (11.20L-2.0L)
w = -13.8atm (convert this to J)
w = -13.8atm * (8.3145 J K^-1 mol^-1)/(0.08206 L atm K^-1 mol^-1)
w= - 1398.2J?
then Delta U = q + w
delta U = -256J + (-1398.2J)
delta U = 1654J?
Is this right?
yes
Your calculation for q and ΔU is correct. However, there is a mistake in the calculation of work (w).
To calculate work, you need to use the formula:
w = -PΔV
where P is the external pressure and ΔV is the change in volume. In this case, the initial volume (V1) is 2.00 L and the final volume (V2) is 11.20 L. Therefore, the change in volume is:
ΔV = V2 - V1 = 11.20 L - 2.00 L = 9.20 L
Now you can calculate the work:
w = -(1.500 atm)(9.20 L)
Note that you do not need to convert atm to J in this case because both units are compatible. Therefore, you do not need to use the conversion factor (0.08206 L atm K^-1 mol^-1).
Using the given values:
w = -13.80 L atm
So, the correct value for work is -13.80 L atm, without converting it to J.
Therefore, the updated answer is:
q = -256 J
w = -13.80 L atm
ΔU = q + w = -256 J + (-13.80 L atm)
Your calculation for q and w are correct, but there is an error in the calculation of ΔU (change in internal energy). Remember that ΔU is the sum of q (heat) and w (work), so you need to add q and w together. In this case, since q is negative (-256 J) and w is also negative (-1398.2 J), when you add them together, the result should be negative as well.
Therefore, the correct calculation for ΔU is:
ΔU = q + w
ΔU = (-256 J) + (-1398.2 J)
ΔU = -1654.2 J
So, the correct value for ΔU is -1654.2 J, not 1654 J.
Note: In this problem, you have correctly calculated w as work done by the gas on the surroundings, which is negative because the gas is expanding against an external pressure. However, be careful with the unit conversion of atm to J. The correct conversion factor is 1 atm = 101.325 J, not (8.3145 J K^-1 mol^-1)/(0.08206 L atm K^-1 mol^-1).