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 gas (H₂ or O₂) will have more grams of molecules hitting a unit area in unit time, we can use the ideal gas equation and some basic principles of gas behavior.

### Step-by-Step Solution:

**Step 1: Understand the Ideal Gas Equation**
The ideal gas equation is given by:
\[ PV = nRT \]
Where:
- \( P \) = Pressure
- \( V \) = Volume
- \( n \) = Number of moles
- \( R \) = Universal gas constant
- \( T \) = Absolute temperature

**Step 2: Relate Mass to Moles**
The number of moles \( n \) can be expressed in terms of mass \( W \) and molar mass \( M \):
\[ n = \frac{W}{M} \]
Thus, we can rewrite the ideal gas equation as:
\[ PV = \frac{W}{M} RT \]
From this, we can express the mass \( W \):
\[ W = \frac{PVM}{RT} \]

**Step 3: Analyze Case I (H₂ at 1 atm and 50 K)**
For H₂:
- \( P = 1 \, \text{atm} \)
- \( V = 1 \, \text{L} \) (unit volume)
- \( M = 2 \, \text{g/mol} \) (molar mass of H₂)
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 50 \, \text{K} \)

Substituting into the equation:
\[ W_{H_2} = \frac{(1 \, \text{atm})(1 \, \text{L})(2 \, \text{g/mol})}{(0.0821 \, \text{L atm/(K mol)})(50 \, \text{K})} \]

Calculating:
\[ W_{H_2} = \frac{2}{4.105} \approx 0.487 \, \text{g} \]

**Step 4: Analyze Case II (O₂ at 2 atm and 200 K)**
For O₂:
- \( P = 2 \, \text{atm} \)
- \( V = 1 \, \text{L} \)
- \( M = 32 \, \text{g/mol} \) (molar mass of O₂)
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 200 \, \text{K} \)

Substituting into the equation:
\[ W_{O_2} = \frac{(2 \, \text{atm})(1 \, \text{L})(32 \, \text{g/mol})}{(0.0821 \, \text{L atm/(K mol)})(200 \, \text{K})} \]

Calculating:
\[ W_{O_2} = \frac{64}{16.42} \approx 3.90 \, \text{g} \]

**Step 5: Compare the Results**
- Mass of H₂ hitting unit area in unit time = \( 0.487 \, \text{g} \)
- Mass of O₂ hitting unit area in unit time = \( 3.90 \, \text{g} \)

**Conclusion:**
Since \( 3.90 \, \text{g} \) (O₂) is greater than \( 0.487 \, \text{g} \) (H₂), more grams of gas molecules will hit a unit area in unit time in the case of O₂.

### Final Answer:
**O₂ at 2 atm and 200 K has more grams of gas molecules hitting a unit area in unit time.**

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