Question

kinetic molecular theory

1. consider a 1.0-l sample of helium gas and a 1.0-l sample of argon gas, both at room temperature and atmospheric pressure.

a. do the atoms in the helium sample have the same average kinetic energy as the atoms in the argon sample?
b. do the atoms in the helium sample have the same average velocity as the atoms in the argon sample?
c. do the argon atoms, because they are more massive, exert a greater pressure on the walls of the container? explain.
d. which gas sample has the faster rate of effusion?
missed this? read section 6.8; watch kcv 6.8

Explanation:

Part (a)

Step1: Recall Kinetic Energy Formula

The average kinetic energy (\(KE_{avg}\)) of gas particles is given by \(KE_{avg}=\frac{3}{2}RT\) (for ideal gases), where \(R\) is the gas constant and \(T\) is the absolute temperature.

Step2: Compare Temperatures

Both helium and argon samples are at room temperature, so their absolute temperatures (\(T\)) are equal.

Step3: Determine Kinetic Energy

Since \(KE_{avg}\) depends only on \(T\) (for ideal gases) and \(T\) is the same for both, the average kinetic energy of helium atoms and argon atoms is the same.

Step1: Recall Velocity Formula

The root - mean - square velocity (\(v_{rms}\)) of gas particles is given by \(v_{rms}=\sqrt{\frac{3RT}{M}}\), where \(M\) is the molar mass of the gas.

Step2: Compare Molar Masses

The molar mass of helium (\(M_{He}\)) is approximately \(4\space g/mol\) and the molar mass of argon (\(M_{Ar}\)) is approximately \(40\space g/mol\). So \(M_{He}

Step3: Determine Velocity

From the formula \(v_{rms}=\sqrt{\frac{3RT}{M}}\), when \(T\) is constant, a smaller molar mass (\(M\)) leads to a larger \(v_{rms}\). Since \(M_{He}

Step1: Recall Ideal Gas Law

The ideal gas law is \(PV = nRT\). We can also think about pressure in terms of molecular collisions. Pressure is the force per unit area due to the collisions of gas particles with the container walls.

Step2: Analyze Conditions

Both samples have the same volume (\(V = 1.0\space L\)), the same temperature (\(T\), room temperature), and the same pressure (atmospheric pressure) initially? Wait, no, the problem says both are at room temperature and atmospheric pressure? Wait, the problem states "a 1.0 - L sample of helium gas and a 1.0 - L sample of argon gas, both at room temperature and atmospheric pressure". From the ideal gas law \(PV=nRT\), since \(P\), \(V\), and \(T\) are the same for both samples, \(n=\frac{PV}{RT}\) is the same for both. The number of moles (\(n\)) is the same. The pressure is determined by the force of collisions per unit area. The average force per collision depends on the momentum change of the particles. The momentum (\(p = mv\)) of argon atoms is higher (since \(m_{Ar}>m_{He}\) and from part (b), even though \(v_{He}>v_{Ar}\), the product \(m_{Ar}v_{Ar}\) and \(m_{He}v_{He}\): let's check with \(v_{rms}=\sqrt{\frac{3RT}{M}}\) and \(p = m\times v_{rms}=m\times\sqrt{\frac{3RT}{M}}=\sqrt{3RTm^{2}/M}=\sqrt{3RTm}\) (since \(M = m\times N_{A}\), but for a single atom, \(m\) is the mass of one atom, and \(M = m\times N_{A}\), so \(m=\frac{M}{N_{A}}\)). Wait, actually, from \(KE_{avg}=\frac{1}{2}mv^{2}=\frac{3}{2}kT\) (where \(k\) is Boltzmann's constant), so \(mv^{2}=3kT\), and momentum \(p = mv\), so \(p=\sqrt{3mkT}\). Since the mass of an argon atom (\(m_{Ar}\)) is greater than the mass of a helium atom (\(m_{He}\)), the momentum of an argon atom is greater than that of a helium atom. However, the number of moles \(n\) is the same for both (from \(PV = nRT\), \(P\), \(V\), \(T\) same, so \(n\) same). The number of particles \(N=nN_{A}\) is the same. The pressure \(P=\frac{1}{3}\frac{N}{V}mv_{rms}^{2}\) (derived from kinetic theory). But we know from \(PV=nRT\) that \(P\) is the same for both (since \(n\), \(V\), \(T\) same). Wait, the problem says "do the argon atoms, because they are more massive, exert a greater pressure on the walls of the container?" But according to the ideal gas law and kinetic theory, at the same \(n\), \(V\), and \(T\), the pressure is the same. The force per collision is greater for argon (because more massive, greater momentum change per collision), but the frequency of collisions (number of collisions per unit time per unit area) is less for argon (since it has a lower velocity). These two effects (greater force per collision and lower collision frequency) cancel out, resulting in the same pressure.

Answer:

Yes, because average kinetic energy of gas particles depends only on temperature (\(KE_{avg}=\frac{3}{2}RT\)), and both are at the same room temperature.

Part (b)