In which of the following two cases more number of grams of gas molecule will hit a unit area in unit time?
(I) `H_2`at 1 atm and 50K
(II) `O_2` at 2atm and 200K

Options

A. Hydrogen
B. Oxygen
C. Same in both the cases
D. Information is incomplete, hence, it can not be predicted.

Answer

To determine which of the two cases results in more grams of gas molecules hitting a unit area in unit time, we will use the ideal gas equation and some basic principles of gas behavior.

### Step-by-Step Solution:

1. **Understand the Problem:**
We need to compare the number of grams of gas molecules hitting a unit area in unit time for two different gases under different conditions.

2. **Identify Given Data:**
- Case I: \( H_2 \) at 1 atm and 50 K
- Case II: \( O_2 \) at 2 atm and 200 K

3. **Use the Ideal Gas Equation:**
The ideal gas equation is given by:
\[ PV = nRT \]
where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = ideal gas constant
- \( T \) = temperature in Kelvin

4. **Calculate the Mass of Gas in Each Case:**
We can rearrange the ideal gas equation to find the mass (\( W \)):
\[ W = \frac{PVM}{RT} \]
where \( M \) is the molar mass of the gas.

5. **Calculate for Case I (Hydrogen, \( H_2 \)):**
- Pressure \( P = 1 \) atm
- Volume \( V = 1 \) L (assuming unit volume)
- Molar mass of \( H_2 = 2 \) g/mol
- Temperature \( T = 50 \) K

Substituting these values:
\[ W_{H_2} = \frac{(1 \text{ atm})(1 \text{ L})(2 \text{ g/mol})}{R \cdot 50 \text{ K}} \]
Simplifying:
\[ W_{H_2} = \frac{2}{50R} = \frac{1}{25R} \]

6. **Calculate for Case II (Oxygen, \( O_2 \)):**
- Pressure \( P = 2 \) atm
- Volume \( V = 1 \) L
- Molar mass of \( O_2 = 32 \) g/mol
- Temperature \( T = 200 \) K

Substituting these values:
\[ W_{O_2} = \frac{(2 \text{ atm})(1 \text{ L})(32 \text{ g/mol})}{R \cdot 200 \text{ K}} \]
Simplifying:
\[ W_{O_2} = \frac{64}{200R} = \frac{32}{100R} = \frac{16}{50R} \]

7. **Comparison of Masses:**
Now we compare \( W_{H_2} \) and \( W_{O_2} \):
\[ W_{H_2} = \frac{1}{25R} \quad \text{and} \quad W_{O_2} = \frac{16}{50R} = \frac{8}{25R} \]

Since \( \frac{8}{25R} > \frac{1}{25R} \), we conclude that:
\[ W_{O_2} > W_{H_2} \]

8. **Final Conclusion:**
More grams of gas molecules will hit a unit area in unit time in Case II (Oxygen, \( O_2 \)).

### Answer:
The answer is \( O_2 \) at 2 atm and 200 K.