# Chemistry Wk 4 Help

Gas Laws and Kinetic Theory of Gases I

Part 1

Shown below are two identical containers labeled A and B. Container A contains a molecule of an ideal gas, and container B contains two molecules of an ideal gas. Both containers are at the same temperature.

a. How do the pressures in the two containers compare? Be sure to explain your answer.

b. Shown below are four different containers (C, D, E, and F), each with the same volume and at the same temperature. How do the pressures of the gases in the containers compare?

c. Container H below has twice the volume of container G. How will the pressure in the containers compare? Explain your reasoning.

d. How will the pressure of containers G and H compare if you add two more gas molecules to container H?

e. Consider containers J and L below. Container L has twice the volume of container J. Container J is at a temperature of 100 K, and container L is at 200 K. How does the pressure in container J compare with that in container L? Include an explanation as part of your answer.

#### 5.28. Gas Laws and Kinetic Theory of Gases II

Consider the box below that contains a single atom of an ideal gas.

a. Assuming that this gas atom is moving, describe how it creates pressure inside the container.

b. Now consider the two containers below, each at the same temperature. If we were to measure the gas pressure in each container, how would they compare? Explain your answer.

c. Consider the two containers below, labeled C and D, one with an ideal gas atom and one with ideal gas molecules, each at the same temperature. The gas molecule has more mass than the gas atom. How do the pressures of the two containers compare? Why?

d. For the containers C and D above, which would have the higher root-mean-square (rms) molecular speed? How does the difference in rms speeds support or contradict the answer you obtained above when you compared the pressures?

e. Consider containers E and F below. How do the pressures in these containers compare? Be sure to explain your answer.

f. Consider containers G and H below, one with a gas atom and one with a gas molecule. As before, the gas molecule has more mass than the gas atom. Explain how the pressures in these containers compare.

g. Now think about the containers below, I and J. How do the pressures in these containers compare? Be sure to justify your answer.

h. Finally, how do the pressures in containers K and L compare?

Part 1.

a) the pressure in container B is higher (double) than the pressure in container A. At the same temperature and volume the general equation of gases states that $P*V= \nu*R*T$ , where $\nu$ is the number of gas moles. Since in the second container there are double the molecules than in the first and the number of molecules in a gas mole is the same for all substances, then its pressure is double.

b) As at point a, the general equation of state for a gas is $P*V= \nu*R*T$ are $nu$ is the number of moles. The volume is the same. The number of molecules per mole is the same regarding the gas. Hence

$\nu _A=\nu _B=\nu _C = (1/2) \nu _D$

$Pressure A= Pressure B= Pressure C = (1/2)Pressure D$

The pressure in D is highest and double from the others

c) $P*V =\nu*R*T$, $T$ is the same, $\nu$ is the same

$V(H) =2V(G)$

$P(H)=1/2*P(G)$

d) $\nu (H) = 2*\nu(G)$.

$V(H) = 2V(G)$

it means

$P(H) = P(G)$

e) $PV=\nu*R*T$

$\nu (L) = \nu (J)$

$T(L)=2T(J)$

$V(L) = 2V(J)$

it means $P(L) = P(J)$

2. Gas laws and kinetic theory

a) The pressure inside a container is created by the collisions of the molecules inside the containers with its wall. By regarding these collisions as elastic then there need to be a force acting on the wall at the collision time to change the total momentum of the molecule (the total momentum after the collision of the molecule needs to be equal in module and having contrary sign to that before the collision).

b) $P*V = \nu*R*T$

$T(B) = T(A)$

$\nu (B) =4\nu (A)$

$V(A) = V(B)$

it means $P(B) = 4P(A)$

c) It does not matter that in the first container there are single atoms and in the second two atoms molecules. Since the number of molecules (or atoms) is the same in a mole of any substance than

$\nu(D) = 4\nu(C)$

$T(D) = T(C)$

$V(D)= V(C)$

it means

$P(D) = 4P(C)$

d) The root mean square speed is

$Vrms = \sqrt{3*R*T/m}$, where $R$ is the gas constant and m is the mass of molecule (or atom) of gas

since $T(D) =T(C)$ and $m(D) =2m(C)$

$Vrms(D) = 1/\sqrt{2} * Vrms(C)$

It does not contradict the answer obtained at point d. Vrms is only a root mean speed.

e) $PV= \nu*R*T$

$V(E) = V(F)$

$T(F) = 2T(E)$

$V(F) = V(E)$

it means

$P(F) =2P(E)$

f) It does not matter the mass of molecule. The number of moles in both container are the same.

$\nu(H) = \nu(G)$

$T(H) = 2T(G)$

$V(H) = V(G)$

it means $P(H) =2P(G)$

g) again it does not matter the mass of molecule. It matter the number of molecule.

$\nu(J) =2\nu(I)$

$T(J) = 1/2 *T(I)$

$V(J) = V(I)$

it means $P(J)=P(I)$

h) $\nu(K) = 3\nu(L)$

$T(K) =2T(L)$

$V(K) =V(L)$

it means $P(K) = 6P(L)$