One gram mole of oxygen at `27^@C` and one atmospheric pressure is enclosed in a vessel. Assuming the molecules to be moving with `v_(rms)`, find the number of collisions per second which the molecules make against one square metre of the vessel wall.

Answer

To solve the problem of finding the number of collisions per second that one gram mole of oxygen molecules makes against one square meter of the vessel wall at 27°C and one atmospheric pressure, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Molar Mass of Oxygen:**
The molar mass of oxygen (O2) is given as 32 g/mol.

2. **Calculate the Mass of One Molecule:**
To find the mass of a single oxygen molecule, we use Avogadro's number (N_A = 6.023 × 10²³ mol⁻¹):
\[ \text{Mass of one molecule} = \frac{\text{Molar mass}}{N_A} = \frac{32 \times 10^{-3} \text{ kg}}{6.023 \times 10^{23}} \approx 5.316 \times 10^{-26} \text{ kg} \]

3. **Convert Temperature to Kelvin:**
The temperature given is 27°C. To convert this to Kelvin:
\[ T = 27 + 273 = 300 \text{ K} \]

4. **Calculate the Root Mean Square Velocity (v_rms):**
The formula for the root mean square velocity is:
\[ v_{rms} = \sqrt{\frac{3kT}{m}} \]
where \( k \) is Boltzmann's constant \( (1.38 \times 10^{-23} \text{ J/K}) \) and \( m \) is the mass of one molecule.
Substituting the values:
\[ v_{rms} = \sqrt{\frac{3 \times (1.38 \times 10^{-23}) \times 300}{5.316 \times 10^{-26}}} \approx 483.35 \text{ m/s} \]

5. **Calculate the Change in Momentum (Δp):**
The change in momentum for one molecule colliding elastically with the wall is:
\[ \Delta p = 2mv_{rms} = 2 \times (5.316 \times 10^{-26}) \times (483.35) \approx 5.14 \times 10^{-23} \text{ kg m/s} \]

6. **Relate Pressure to Number of Collisions:**
The pressure \( P \) is given as 1 atm, which we convert to Pascals:
\[ P = 1 \text{ atm} = 1.01325 \times 10^{5} \text{ Pa} \]
The pressure can also be expressed in terms of the number of collisions per second \( n \):
\[ P = n \cdot \Delta p \]

7. **Calculate the Number of Collisions (n):**
Rearranging the equation gives:
\[ n = \frac{P}{\Delta p} = \frac{1.01325 \times 10^{5}}{5.14 \times 10^{-23}} \approx 1.96 \times 10^{27} \text{ collisions/s} \]
Rounding this gives:
\[ n \approx 2 \times 10^{27} \text{ collisions/s} \]

### Final Answer:
The number of collisions per second that the molecules make against one square meter of the vessel wall is approximately \( 2 \times 10^{27} \).