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?
Your approach is almost correct, but there are a couple of mistakes in your calculations.
First, let's calculate the work done by the gas:
w = -PΔV
w = -(1.5 atm) * (11.20 L - 2.00 L)
w = -13.8 atm * L
To convert atm * L to joules, you need to multiply by the conversion factor 101.325 J/atm.
w = -13.8 atm * L * (101.325 J/atm)
w = -1400 J
So the correct value for work (w) is -1400 J, not -1398.2 J.
Next, let's calculate the change in internal energy (ΔU):
ΔU = q + w
ΔU = -256 J + (-1400 J)
ΔU = -1656 J
So the correct value for the change in internal energy (ΔU) is -1656 J, rather than 1654 J.
To summarize:
q = -256 J
w = -1400 J
ΔU = -1656 J