Yes. They obviously want you to assume that the expansion is reversible. Your equation is correct if W represents the work done by the surroundings on the system.krootox217 said:
This one?
No. ΔUm is not equal to zero.krootox217 said:
You don't need to ask me this. Why don't you solve the same the same problem using the ideal gas law and see how the numbers compare? Incidentally, why did you need wolframalpha to do the integration for you? Why didn't you do the integration yourself?krootox217 said:
Sure. Wouldn't you have expected that?krootox217 said:
Q and W both refer to different forms of energy in thermodynamics. Q stands for heat, while W stands for work. In the case of Van-der-Waals gases, Q refers to the heat absorbed or released during a process, while W refers to the work done on or by the gas.
In thermodynamics, the first law states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. This means that for Van-der-Waals gases, Q and W are related by the equation ΔU = Q - W.
Q and W play important roles in understanding the behavior of Van-der-Waals gases. The heat absorbed or released by the gas can affect its temperature, while the work done on or by the gas can change its volume. This can ultimately impact the overall behavior and properties of the gas.
As mentioned before, Q and W are directly related to the internal energy of a system. An increase in Q will result in an increase in internal energy, while an increase in W will lead to a decrease in internal energy. This means that changes in Q and W can directly impact the energy state of Van-der-Waals gases.
Q and W can be manipulated by changing the conditions of the gas, such as pressure, temperature, and volume. For example, increasing the pressure on a gas will result in more work being done by the gas, while increasing the temperature will lead to more heat being absorbed by the gas. This can ultimately affect the behavior of Van-der-Waals gases.