Two vessels "A" and "B" are of the same size and are at same temperature.A contains 1g of hydrogen and "B" contains "1g" of oxygen.`P_(A)` and `P_(B` are the pressures of the "],[" gases in "A" and "B" respectively,then `(P_(A))/(P_(B))` is:

Options

A. 16
B. 8
C. 32
D. 4

Answer

To solve the problem, we will use the Ideal Gas Law, which states that:

\[ PV = nRT \]

Where:
- \( P \) = pressure of the gas
- \( V \) = volume of the gas
- \( n \) = number of moles of the gas
- \( R \) = universal gas constant
- \( T \) = temperature of the gas

### Step 1: Determine the number of moles of each gas

1. **For Hydrogen (H₂)**:
- The molar mass of hydrogen (H₂) is approximately 2 g/mol.
- Given mass of hydrogen = 1 g.
- Number of moles of hydrogen (\( n_A \)):
\[ n_A = \frac{\text{mass}}{\text{molar mass}} = \frac{1 \text{ g}}{2 \text{ g/mol}} = 0.5 \text{ moles} \]

2. **For Oxygen (O₂)**:
- The molar mass of oxygen (O₂) is approximately 32 g/mol.
- Given mass of oxygen = 1 g.
- Number of moles of oxygen (\( n_B \)):
\[ n_B = \frac{\text{mass}}{\text{molar mass}} = \frac{1 \text{ g}}{32 \text{ g/mol}} = 0.03125 \text{ moles} \]

### Step 2: Apply the Ideal Gas Law to both vessels

- For vessel A (Hydrogen):
\[ P_A V = n_A RT \implies P_A = \frac{n_A RT}{V} \]

- For vessel B (Oxygen):
\[ P_B V = n_B RT \implies P_B = \frac{n_B RT}{V} \]

### Step 3: Find the ratio of pressures \( \frac{P_A}{P_B} \)

Now we can find the ratio of the pressures:
\[ \frac{P_A}{P_B} = \frac{\frac{n_A RT}{V}}{\frac{n_B RT}{V}} = \frac{n_A}{n_B} \]

### Step 4: Substitute the values of \( n_A \) and \( n_B \)

Substituting the values we calculated:
\[ \frac{P_A}{P_B} = \frac{0.5}{0.03125} = 16 \]

### Conclusion

Thus, the ratio of the pressures \( \frac{P_A}{P_B} \) is:

\[ \frac{P_A}{P_B} = 16 \]

### Final Answer:

\[ \frac{P_A}{P_B} = 16 \]

---