To solve the problem step by step, we will break it down into two parts as given in the question.
### Part (a): Number of Collisions per Second
1. **Understand the Given Data:**
- 1 g mole of oxygen (O₂)
- Temperature (T) = 27°C = 300 K (in Kelvin)
- Pressure (P) = 1 atm = \(1 \times 10^5\) Pa
- Area (A) = 1 m²
2. **Calculate the Force Exerted on the Wall:**
\[
\text{Force} (F) = \text{Pressure} (P) \times \text{Area} (A) = 1 \times 10^5 \, \text{Pa} \times 1 \, \text{m}^2 = 1 \times 10^5 \, \text{N}
\]
3. **Relate Force to Change in Momentum:**
The change in momentum per second due to collisions can be expressed as:
\[
F = \frac{\Delta P}{\Delta t}
\]
where \(\Delta P\) is the change in momentum and \(\Delta t\) is the time interval (1 second for our calculation).
4. **Change in Momentum for Collisions:**
For each molecule colliding with the wall, the change in momentum is given by:
\[
\Delta P = 2mv
\]
where \(m\) is the mass of one molecule and \(v\) is the velocity of the molecule.
5. **Calculate the Number of Molecules (n):**
The number of collisions per second (n) can be expressed as:
\[
n = \frac{F}{2mv}
\]
6. **Calculate the RMS Velocity (v_rms):**
The RMS velocity for an ideal gas is given by:
\[
v_{rms} = \sqrt{\frac{3RT}{M}}
\]
where:
- \(R = 8.314 \, \text{J/(mol K)}\)
- \(T = 300 \, \text{K}\)
- \(M = 32 \, \text{g/mol} = 0.032 \, \text{kg/mol}\)
Substituting the values:
\[
v_{rms} = \sqrt{\frac{3 \times 8.314 \times 300}{0.032}} \approx 483.4 \, \text{m/s}
\]
7. **Calculate the Mass of One Molecule:**
The mass of one molecule of oxygen:
\[
m = \frac{M}{N_A} = \frac{0.032 \, \text{kg/mol}}{6.022 \times 10^{23} \, \text{molecules/mol}} \approx 5.31 \times 10^{-26} \, \text{kg}
\]
8. **Substitute Values into the Collision Formula:**
Now substituting \(F\), \(m\), and \(v_{rms}\) into the equation for \(n\):
\[
n = \frac{1 \times 10^5}{2 \times (5.31 \times 10^{-26}) \times (483.4)} \approx 1.97 \times 10^{27} \, \text{collisions/second}
\]
### Part (b): Calculate the Speed \(v_0\)
1. **Understand the Process:**
The vessel is thermally insulated and moves with a constant speed \(v_0\). When it stops, the temperature of the gas rises by \(1^\circ C\).
2. **Use Conservation of Energy:**
The change in kinetic energy as the vessel stops results in a change in internal energy of the gas:
\[
\Delta KE = \Delta U
\]
where \(\Delta KE = \frac{1}{2} mv_0^2\) and \(\Delta U = nC_v \Delta T\).
3. **Calculate \(C_v\) for Oxygen:**
For a diatomic gas like oxygen:
\[
C_v = \frac{5}{2} R = \frac{5}{2} \times 8.314 \approx 20.79 \, \text{J/(mol K)}
\]
4. **Substituting Values:**
Since \(n = 1 \, \text{mol}\) and \(\Delta T = 1 \, \text{K}\):
\[
\Delta U = 1 \times 20.79 \times 1 = 20.79 \, \text{J}
\]
5. **Set Up the Equation:**
\[
\frac{1}{2} mv_0^2 = 20.79
\]
6. **Substituting for Mass:**
\[
m = 0.032 \, \text{kg}
\]
So,
\[
\frac{1}{2} (0.032)v_0^2 = 20.79
\]
7. **Solve for \(v_0\):**
\[
v_0^2 = \frac{2 \times 20.79}{0.032} \approx 1305.94
\]
\[
v_0 \approx \sqrt{1305.94} \approx 36.1 \, \text{m/s}
\]
### Final Answers:
- (a) Number of collisions per second: \(n \approx 1.97 \times 10^{27} \, \text{collisions/second}\)
- (b) Speed \(v_0 \approx 36 \, \text{m/s}\)
