The volume of an ideal gas contracts from 5.68 L to 2.35 L as the result of a constant applied pressure of 1.48 atm. The system releases 733 J of heat during this process. Calculate deltaU (internal energy) for this process. Express answer in Joules.
Please help, thank you very much.
To calculate the change in internal energy (ΔU) for this process, you can 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.
The formula for calculating ΔU is:
ΔU = Q - W
Where:
ΔU = change in internal energy
Q = heat added to the system
W = work done by the system
In this case, we are given the heat released by the system (Q = -733 J), but we need to calculate the work done by the system.
The work done by a system can be calculated using the formula:
W = -PΔV
Where:
W = work done by the system
P = pressure
ΔV = change in volume
We are given that the pressure is 1.48 atm, the initial volume is 5.68 L, and the final volume is 2.35 L.
Substituting the given values into the formula, we get:
ΔV = V_final - V_initial
= 2.35 L - 5.68 L
= -3.33 L
W = -PΔV
= -(1.48 atm)(-3.33 L)
= 4.9324 L·atm
Now we have the values for Q and W, so we can calculate ΔU:
ΔU = Q - W
= -733 J - 4.9324 L·atm
To express the answer in Joules, we need to convert the unit of work from L·atm to Joules. Recall that 1 L·atm = 101.325 J.
W in Joules = (4.9324 L·atm)(101.325 J / 1 L·atm)
= 499.81153 J
ΔU = -733 J - 499.81153 J
= -1232.8115 J
Therefore, the change in internal energy (ΔU) for this process is approximately -1232.8115 J.