What is the change in internal energy of a system (ΔU) if 153 J of heat is released by the system and 250J of work is done on the system by the surroundings?
What I did:
ΔU = -153 - (+250)
ΔU = - 403J
I also have a question about the Van der Waals constants, if b is zero and the gas is under high pressure would the volume of the gas would be exactly the same as the Ideal Gas Law?
I think that it would be exactly the same as the Ideal Gas Law because Van der Waals constants for is V-nb, if b is zero then V will get subtract by nothing V-0, but I'm not sure.
How about low pressure how would they affect the outcome? Finally if b had a value how would it affect the gas?
first question on dU. It was my impression that dU = dQ + dW. Your signs are right. Heat released by the system is -153 J and work don ON the system is +250 J
I'll look at part 2 later.
With regard to the b = 0 then you are right that V-0 = V; however, I don't thin P would be the same as the ideal gas law BECAUSE you still have the a factor involved. P would be the same as the ideal gas law, it seems to me, when BOTH a and b are 0 so that the van der Waals equation is reduced to the ideal gas law equation.
Then how would "a" affect the volume of gas? I'm confused on that part?
Looking at the a and b values for real gases in the Van der Waals equations, the ones I looked at were positive values; i.e., no negative values in the half dozen or so I saw. So if b = 0 and we include the constant for a, no matter what it is, then P + constant will be higher total P value and that should cause a lower volume to be calculated.
I want to point out that the Van der Waals equation is not a "fix it all for all conditions". It is a decent model for decent conditions of real gases but I copied this from Wikipedia. "This equation approximates the behavior of real fluids above their critical temperatures and is qualitatively reasonable for their liquid and low-pressure gaseous states at low temperatures. However, near the phase transitions between gas and liquid, in the range of p, V, and T where the liquid phase and the gas phase are in equilibrium, the Van der Waals equation fails to accurately model observed experimental behavior. In particular, p is a constant function of V at given temperatures in these regions. As such, the Van der Waals model is not useful for calculations intended to predict real behavior in regions near critical points. Corrections to address these predictive deficiencies include the equal area rule and the principle of corresponding states."
Hope this helps with the confusion.
To calculate the change in internal energy (ΔU) of a system, you need to use the First Law of Thermodynamics, which states that ΔU = Q - W, where Q represents heat and W represents work.
In your case, the system releases 153 J of heat (Q = -153 J) and 250 J of work is done on the system by the surroundings (W = +250 J). To calculate ΔU, you substitute these values into the formula:
ΔU = Q - W
ΔU = -153 J - (+250 J)
ΔU = -403 J
Therefore, the change in internal energy of the system is -403 J.
Now let's move on to the question about Van der Waals constants. In the Van der Waals equation, which is an improvement over the Ideal Gas Law, there are two constants: a and b.
If b is zero, as you mentioned, the volume term in the Van der Waals equation (V - nb) simplifies to just V, which is the same as the Ideal Gas Law. So, under high pressure, when b is zero, the volume of the gas will be exactly the same as predicted by the Ideal Gas Law.
However, at low pressures, the presence of the b constant becomes more significant. The b constant represents the volume of the gas particles themselves, and at low pressures, the volume occupied by the gas particles becomes a more substantial fraction of the total volume. In such cases, the Van der Waals equation provides a better prediction for the behavior of the gas compared to the Ideal Gas Law.
If b has a finite value, it affects the gas by accounting for the volume occupied by the gas particles. The b term reduces the overall volume available to the gas molecules, leading to a slight decrease in the predicted volume of the gas compared to that predicted by the Ideal Gas Law.
In summary, the Van der Waals constants, particularly the value of b, account for the volume occupied by gas particles and become more significant at low pressures, providing a more accurate description of real gases compared to the Ideal Gas Law.