One mole of an ideal gas is expanded from a volume of 1.00 atm. How much work (in joules) is performed on the surroundings? Ignore sig figs for this problem. (T=300 K; 1 L*atm=101.3 J)
Volume of 1 atm? I might believe 1 L but not 1 atm for volume.
To calculate the work performed on the surroundings, we can use the formula:
Work = -P * ΔV
where:
- Work is the work performed on the surroundings in joules (J)
- P is the pressure in atmospheres (atm)
- ΔV is the change in volume in liters (L)
In this case, we have a one-mole ideal gas expanded from a volume of 1.00 L at a pressure of 1.00 atm. We need to determine the change in volume, ΔV.
Since we have an ideal gas, we can use the ideal gas law equation:
PV = nRT
where:
- P is the pressure in atmospheres (atm)
- V is the volume in liters (L)
- n is the number of moles of the gas
- R is the ideal gas constant (0.0821 L*atm/(K*mol))
- T is the temperature in Kelvin (K)
Solving the ideal gas law equation for V, we get:
V = (nRT) / P
Substituting the known values into the equation, we have:
V = (1 mol * 0.0821 L*atm/(K*mol) * 300 K) / (1.00 atm)
V ≈ 24.63 L
Now we can calculate the change in volume, ΔV:
ΔV = final volume - initial volume
ΔV = 24.63 L - 1.00 L
ΔV = 23.63 L
Finally, we can calculate the work using the formula mentioned earlier:
Work = -P * ΔV
Work = -(1.00 atm) * (23.63 L)
Work ≈ -23.63 L*atm
Since we need the work in joules, we can convert from L*atm to joules using the given conversion factor:
1 L*atm = 101.3 J
Work ≈ -23.63 L*atm * (101.3 J / 1 L*atm)
Work ≈ -2391.619 J
Therefore, the work performed on the surroundings is approximately -2391.619 joules. The negative sign indicates work done on the surroundings.