We have a container of a hot ideal monatomic gas. The volume of the container is

Question

We have a container of a hot ideal monatomic gas. The volume of the container is 25 liters. The temperature of the gas is 409 degrees C and its pressure is 0.858x10^5 Pa. We allow the gas to cool down to room temperature, which at the time is 21 degrees C. We do not allow the volume of the gas to change. (a) Find the final pressure (Pa) of the gas. (b) Find the amount of heat (J) that passes from the gas to its surroundings as it cooled. (Give a positive value) (c) Find the change of internal energy (J) of the gas. Be sure to include the correct sign on the answer.

Explanation / Answer

a.) Okay so for part a we first want to convert all of our data into MKS units used in the ideal gas equation (this means volumes are in meters cubed, pressures are in pascals, and temperatures are in Kelvin). If the measurements are not in the correct units the equation will not work.

So lets first convert liters to meters cubed:

(25 L)(1 L/1000 m^3) = .025 m^3

Now lets convert temperatures:

T initial = 273 + 309
T initial = 582 K

T final = 273 + 21
T final = 294 K

Now it’s time to use the ideal gas equation, PV = NKT.

In the initial situation, we can solve for the quantity NK very easily:

PV/T = NK

[(0.858x10^5)(.025)]/582 = NK

Since the gas is in a closed container and no gas can escape, N the number of molecules does not change. Furthermore, since K is a constant, it does not change either.

In the final situation, we know V (because it does not change as stated in the problem), NK, and T. Therefore, we can solve for P the pressure in the final situation.

P(.025) = ([(0.858x10^5)(.025)]/582)(294)

Notice that .025 cancels on both sides leaving:

P =([(0.858x10^5)(294)]/582

P = 4.33 x 10^4 Pa

b.) To solve part b of the problem we will use the first law of thermodynamics which states that
dU = dQ + dW
Since the thermodynamic process mentioned is isochoric (no volume change) the dW term is 0.

Therefore, for the situation in question dU=dQ.

Recall that dU = (3/2)NKdT.

Thus dQ = (3/2)NKdT

We found NK in part a of the problem, and dT is simply the final temperature minus the initial temperature.

dT = (294-582)
dT = -288 K

dQ = (3/2)([(0.858x10^5)(.025)]/582)(-288)

dQ = -1592.2 Joules

Note that the dQ is negative which makes sense because heat must be lost in order for the temperature to decrease if the volume stays constant.

Therefore, the amount of heat lost to the surroundings is 1592.2 Joules!

c.) Part c has actually already been answered! Remember how dU=dQ? Well we found dQ in the last part so we know dU.

dU = -1592.2 Joules

dU is negative because there is a drop in internal energy. The drop in internal energy occurs because the temperature decresed.

Hope this helped!

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