1 gram `H_(2)` and 8g `O_(2)` were taken in a 10 liter vessel at 300 K. The partial pressure exerted by `O_(2)` will be

Options

A. 1.84 atm
B. 0.88 atm
C. 1.03 atm
D. 0.61 atm

Answer

To find the partial pressure exerted by \( O_2 \) in a 10-liter vessel at 300 K, we can use the ideal gas law and the concept of partial pressure. Here’s a step-by-step solution:

### Step 1: Calculate the number of moles of \( O_2 \)

The molar mass of \( O_2 \) is approximately 32 g/mol. Given that we have 8 g of \( O_2 \):

\[ \text{Number of moles of } O_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{8 \, \text{g}}{32 \, \text{g/mol}} = 0.25 \, \text{mol} \]

### Step 2: Use the ideal gas law to find the partial pressure of \( O_2 \)

The ideal gas law is given by:

\[ PV = nRT \]

Where:
- \( P \) is the pressure,
- \( V \) is the volume (10 L),
- \( n \) is the number of moles (0.25 mol),
- \( R \) is the ideal gas constant (0.0821 L·atm/(K·mol)),
- \( T \) is the temperature in Kelvin (300 K).

Rearranging the ideal gas law to solve for pressure \( P \):

\[ P = \frac{nRT}{V} \]

### Step 3: Substitute the values into the equation

Substituting the known values into the equation:

\[ P_{O_2} = \frac{(0.25 \, \text{mol}) \times (0.0821 \, \text{L·atm/(K·mol)}) \times (300 \, \text{K})}{10 \, \text{L}} \]

### Step 4: Calculate the pressure

Now, performing the calculation:

\[ P_{O_2} = \frac{0.25 \times 0.0821 \times 300}{10} \]

Calculating the numerator:

\[ 0.25 \times 0.0821 \times 300 = 6.1575 \]

Now divide by the volume:

\[ P_{O_2} = \frac{6.1575}{10} = 0.61575 \, \text{atm} \]

Rounding to two decimal places, we get:

\[ P_{O_2} \approx 0.62 \, \text{atm} \]

### Final Answer

The partial pressure exerted by \( O_2 \) is approximately **0.62 atm**.

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