Number of molecules in lO0 cc of a gas at I 47. N.T.P. is ___ . if mass of each molecule is `4.556xx10^(-25)`kg and R.M.S speed is 350m/s.

Options

A. `5.4xx10^(19)`
B. `5.4xx10^(21)`
C. `5.4xx10^(18)`
D. `5.4xx10^(23)`

Answer

To find the number of molecules in 100 cc of a gas at NTP, we can use the kinetic theory of gases and the relationship between pressure, density, and RMS speed. Here’s a step-by-step solution:

### Step 1: Understand the Given Data
- Volume (V) = 100 cc = 100 x 10^(-6) m³ = 10^(-4) m³
- Mass of each molecule (m) = 4.556 x 10^(-25) kg
- RMS speed (v_rms) = 350 m/s
- Pressure at NTP (P) = 10^5 Pa

### Step 2: Use the Formula for Pressure
According to the kinetic theory of gases, the pressure (P) can be expressed as:
\[ P = \frac{1}{3} \rho v_{rms}^2 \]
where \(\rho\) is the density of the gas.

### Step 3: Calculate Density
Density (\(\rho\)) can be expressed in terms of the number of molecules (N) and the mass of each molecule (m):
\[ \rho = \frac{N \cdot m}{V} \]

### Step 4: Substitute Density into the Pressure Equation
Substituting the expression for density into the pressure equation gives:
\[ P = \frac{1}{3} \left(\frac{N \cdot m}{V}\right) v_{rms}^2 \]

### Step 5: Rearrange to Solve for N
Rearranging the equation to solve for N:
\[ N = \frac{3PV}{m v_{rms}^2} \]

### Step 6: Substitute the Values
Now, substitute the known values into the equation:
- P = \(10^5\) Pa
- V = \(10^{-4}\) m³
- m = \(4.556 \times 10^{-25}\) kg
- \(v_{rms} = 350\) m/s

So,
\[ N = \frac{3 \times (10^5) \times (10^{-4})}{(4.556 \times 10^{-25}) \times (350^2)} \]

### Step 7: Calculate the Denominator
First, calculate \(350^2\):
\[ 350^2 = 122500 \]

Now, calculate the denominator:
\[ (4.556 \times 10^{-25}) \times 122500 = 5.5707 \times 10^{-20} \]

### Step 8: Calculate N
Now substitute back into the equation for N:
\[ N = \frac{3 \times 10^5 \times 10^{-4}}{5.5707 \times 10^{-20}} \]
\[ N = \frac{3 \times 10^{1}}{5.5707 \times 10^{-20}} \]
\[ N = \frac{3}{5.5707} \times 10^{21} \]
\[ N \approx 5.39 \times 10^{21} \]

### Step 9: Round the Result
Rounding gives:
\[ N \approx 5.4 \times 10^{21} \]

### Final Answer
The number of molecules in 100 cc of gas at NTP is approximately:
\[ \boxed{5.4 \times 10^{21}} \]