Sir I first case we keep v and t constant then how pressure decrease for z=0.9 and pressure increases for Z=1.1 my question is you are adjusting moles in the container for first case removing mole lead to attractive force for second case adding moles
Good question. Let us assume that we have two containers at the same T and V with the same moles of gas. Thus, n, V, and T are constant. If the gases were perfect (ideal, Z=1), then the pressures would be the same. This is because everything in the perfect gas equation (pV=nRT) is constant). But for a real gas pV=ZnRT. So if n, V, T are constant, then the pressure will change if Z is different. If Z=0.9 for one container, attractive forces are dominant. The attractive forces mean that the gas is more attracted to itself and won't collide with the container as much, hence lower pressure.
If Z=0.1 for the other container, repulsice forces are dominant. The repulsive forces mean that the gas is crowded or repels itself and will collide with the conatainer more often, hence greater pressure. Z can change if you change V, n, or T. But Z is also dependent of the type of gas. Does that help?
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Best video ever
Sir I first case we keep v and t constant then how pressure decrease for z=0.9 and pressure increases for Z=1.1 my question is you are adjusting moles in the container for first case removing mole lead to attractive force for second case adding moles
Good question. Let us assume that we have two containers at the same T and V with the same moles of gas. Thus, n, V, and T are constant.
If the gases were perfect (ideal, Z=1), then the pressures would be the same. This is because everything in the perfect gas equation (pV=nRT) is constant).
But for a real gas pV=ZnRT. So if n, V, T are constant, then the pressure will change if Z is different.
If Z=0.9 for one container, attractive forces are dominant. The attractive forces mean that the gas is more attracted to itself and won't collide with the container as much, hence lower pressure.
If Z=0.1 for the other container, repulsice forces are dominant. The repulsive forces mean that the gas is crowded or repels itself and will collide with the conatainer more often, hence greater pressure.
Z can change if you change V, n, or T. But Z is also dependent of the type of gas. Does that help?
@@brainynerdtutor1626 In the part " If Z=0.1" do you mean "Z=1.1"?
@@gj4792yes i’m sure that’s what he means
"if you want to surprise someone at a party" lol
Hope you have a nice life.
You are a lil bit fast 😢