Suppose 2500 J of heat are added to 4.1 mol of argon gas at a constant pressure of 120 kPa. (Assume that the argon can be treated as an ideal monatomic gas.) Find the change in internal energy.
I started using U = N (3/2 kB T). But I have too many unknowns. The only other equation that our teacher gave us is Won + Qon = deltaU (Work on system + Heat on system = Change in internal energy). Please help!
Qon = deltaU - Won =
deltaU + Wsystem,
where Wsystem is the work performed by the system (in this case the argon gas). Because the pressure is constant, the work performed is:
P deltaV = Delta (PV) = (using ideal gas law) Delta (N kB T)
This means that:
Qon = deltaU + Wsystem =
5/2 N kB delta T =
(5/2)(2/3)(3/2)N kB delta T =
(5/2)(2/3) delta U = 5/3 delta U
And it follows that:
delta U = 3/5 Qon = 1500 J
To find the change in internal energy, we can use the equation ΔU = Qon - Won, where Qon is the heat added to the system and Won is the work performed on the system.
Since the pressure is constant, the work performed by the system is given by Wsystem = PΔV, where P is the pressure and ΔV is the change in volume. Using the ideal gas law, PV = NkBT, where P is the pressure, V is the volume, N is the number of moles, kB is the Boltzmann constant, and T is the temperature in Kelvin.
We can rearrange the ideal gas law to solve for ΔV:
ΔV = (NkBΔT) / P
Substituting this into the equation for work, we have:
Wsystem = PΔV = P(NkBΔT) / P = NkBΔT
Now, we can substitute this expression for Wsystem into the equation for ΔU:
ΔU = Qon - Won = Qon - Wsystem
Since the argon gas can be treated as an ideal monatomic gas, the internal energy can be expressed as ΔU = (5/2)NkBΔT.
Now we can substitute ΔU = (5/2)NkBΔT and Wsystem = NkBΔT back into the equation:
ΔU = Qon - Wsystem
(5/2)NkBΔT = Qon - NkBΔT
Simplifying this equation, we have:
ΔU = Qon - (5/2)NkBΔT
ΔU = Qon - (5/2)(2/3)(3/2)NkBΔT
ΔU = Qon - (5/3)ΔU
Now, we can solve for ΔU:
ΔU + (5/3)ΔU = Qon
(8/3)ΔU = Qon
ΔU = (3/8)Qon
Substituting the given value of Qon (2500 J), we have:
ΔU = (3/8)(2500 J)
ΔU = 937.5 J
Therefore, the change in internal energy of the argon gas is 937.5 J.