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 ΔU (change in internal energy) for this process, we need to use the first law of thermodynamics, which states that ΔU = q - w, where ΔU is the change in internal energy, q is the heat added or released, and w is the work done on or by the system.
In this case, we are given that q (heat released) is equal to 733 J. To calculate the work done, we can use the equation w = -PΔV, where P is the pressure and ΔV is the change in volume.
Given:
Initial volume (V1) = 5.68 L
Final volume (V2) = 2.35 L
Applied pressure (P) = 1.48 atm
First, we need to convert the pressure from atm to SI units (Pascal):
1 atm = 101,325 Pa
So, P = 1.48 atm * 101,325 Pa/atm = 149,706.6 Pa
Next, we can calculate the change in volume (ΔV) using the equation ΔV = V2 - V1:
ΔV = 2.35 L - 5.68 L = -3.33 L
Since the volume decreased, the work done on the system is positive, so ΔV should be negative:
ΔV = -3.33 L = -3.33 × 10^(-3) m^3
Now, we can calculate the work done (w) using the equation w = -PΔV:
w = - 149,706.6 Pa * (-3.33 × 10^(-3) m^3)
w = 498.162 J (rounded to three decimal places)
Finally, we can calculate the change in internal energy (ΔU) using the equation ΔU = q - w:
ΔU = 733 J - 498.162 J
ΔU ≈ 234.838 J
Therefore, the change in internal energy (ΔU) for this process is approximately 234.838 J.