One gram mole `NO_2 "at" 47^@ C` and 2 atm pressure in kept in a vessel. Assuming the molecules to be moving with (rms) velocity. Find the number of collisions per which the molecules make with one square metre area of the vessel wall.

Explanation

To solve the problem of finding the number of collisions per second that 1 gram mole of NO2 at 47°C and 2 atm pressure makes with one square meter area of the vessel wall, we will follow these steps:

### Step 1: Convert Temperature to Kelvin
We need to convert the given temperature from Celsius to Kelvin.
\[ T(K) = T(°C) + 273 = 47 + 273 = 320 \, K \]

### Step 2: Calculate the RMS Velocity
The formula for the root mean square (RMS) velocity (\(v_{rms}\)) of gas molecules is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
Where:
- \(R = 8.314 \, \text{J/(mol K)}\) (universal gas constant)
- \(T = 320 \, K\)
- \(M = 46 \times 10^{-3} \, \text{kg/mol}\) (molar mass of NO2)

Substituting the values:
\[ v_{rms} = \sqrt{\frac{3 \times 8.314 \times 320}{46 \times 10^{-3}}} \]
Calculating this gives:
\[ v_{rms} \approx 423 \, \text{m/s} \]

### Step 3: Calculate the Mass of One Molecule
The mass of one molecule of NO2 can be calculated using:
\[ m = \frac{M}{N_A} \]
Where \(N_A = 6.022 \times 10^{23} \, \text{molecules/mol}\) (Avogadro's number).
Substituting the values:
\[ m = \frac{46 \times 10^{-3}}{6.022 \times 10^{23}} \approx 7.64 \times 10^{-26} \, \text{kg} \]

### Step 4: Use the Pressure Equation
The pressure \(P\) of the gas can be expressed as:
\[ P = \frac{F}{A} = \frac{\Delta p}{\Delta t \cdot A} \]
Where \(\Delta p\) is the change in momentum and \(\Delta t\) is the time interval. The change in momentum for a molecule colliding with the wall is:
\[ \Delta p = 2m v_{rms} \]
Thus, the pressure can be rewritten as:
\[ P = \frac{2m v_{rms} n}{A} \]
Where \(n\) is the number of molecules colliding with the wall per second per square meter.

### Step 5: Rearrange to Find \(n\)
Rearranging the equation to solve for \(n\):
\[ n = \frac{P \cdot A}{2m v_{rms}} \]
Substituting \(P = 2 \, \text{atm} = 2 \times 1.013 \times 10^5 \, \text{Pa}\) and \(A = 1 \, \text{m}^2\):
\[ n = \frac{(2 \times 1.013 \times 10^5) \cdot 1}{2 \cdot (7.64 \times 10^{-26}) \cdot (423)} \]
Calculating this gives:
\[ n \approx 3.1 \times 10^{27} \, \text{collisions/m}^2/s \]

### Final Answer
The number of collisions per second that the molecules make with one square meter area of the vessel wall is approximately:
\[ \boxed{3.1 \times 10^{27}} \]