A sample of gas is compressed from 13.57 L to 3.93 L using a constant external pressure of 2.91 atm. At the same time, 1,759 J of heat flows into the surroundings. What is the internal energy change for the gas?
Isn't this just dE = q + w.
They give you q. s is -p*dV
To determine the internal energy change for the gas, we need to use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (Q) minus the work done by the system (W), i.e., ΔU = Q - W.
In this case, we are given the heat flow (Q) into the surroundings, but we need to find the work done by the system (W). We can calculate the work done using the formula W = -PΔV, where P is the constant external pressure and ΔV is the change in volume.
Given:
Initial volume (V1) = 13.57 L
Final volume (V2) = 3.93 L
Constant external pressure (P) = 2.91 atm
Heat flow (Q) = 1759 J
First, we need to convert the pressure from atm to Pa (using the conversion factor: 1 atm = 101,325 Pa) and the volumes from liters to cubic meters (using the conversion factor: 1 L = 0.001 m^3).
P = 2.91 atm × 101,325 Pa/atm = 295,005.75 Pa
V1 = 13.57 L × 0.001 m^3/L = 0.01357 m^3
V2 = 3.93 L × 0.001 m^3/L = 0.00393 m^3
Now, we can calculate the work done by the system:
W = -PΔV = -295,005.75 Pa × (0.00393 m^3 - 0.01357 m^3)
W = 295,005.75 Pa × (-0.00964 m^3)
W = -2844.45 J
Finally, we can substitute the values of Q and W into the first law of thermodynamics equation to find the internal energy change (ΔU):
ΔU = Q - W
ΔU = 1759 J - (-2844.45 J)
ΔU = 4603.45 J
Therefore, the internal energy change for the gas is 4603.45 J.