A sample of a given mass of gas at a constant temperature occupies `95 cm^(3)` under a pressure of `9.962 xx 10^(4) Nm^(-2)`. At the same temperature its volume at a pressure of 10.13 xx 10^(4) Nm^(-2)` is

Options

A. `190 cm^(3)`
B. `93 cm^(3)`
C. `46.5 cm^(3)`
D. `47.5 cm^(3)`

Explanation

To solve the problem, we will use Boyle's Law, which states that for a given mass of gas at constant temperature, the product of pressure and volume is constant. This can be expressed mathematically as:

\[ P_1 V_1 = P_2 V_2 \]

Where:
- \( P_1 \) = initial pressure
- \( V_1 \) = initial volume
- \( P_2 \) = final pressure
- \( V_2 \) = final volume (which we need to find)

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial volume, \( V_1 = 95 \, \text{cm}^3 \)
- Initial pressure, \( P_1 = 9.962 \times 10^4 \, \text{N/m}^2 \)
- Final pressure, \( P_2 = 10.13 \times 10^4 \, \text{N/m}^2 \)

2. **Apply Boyle's Law:**
Using the formula \( P_1 V_1 = P_2 V_2 \), we can rearrange it to solve for \( V_2 \):
\[ V_2 = \frac{P_1 V_1}{P_2} \]

3. **Substitute the Known Values:**
Substitute \( P_1 \), \( V_1 \), and \( P_2 \) into the equation:
\[ V_2 = \frac{(9.962 \times 10^4) \times 95}{10.13 \times 10^4} \]

4. **Calculate the Numerator:**
\[ (9.962 \times 10^4) \times 95 = 9473.9 \times 10^4 \]

5. **Calculate the Denominator:**
\[ 10.13 \times 10^4 = 10.13 \times 10^4 \]

6. **Perform the Division:**
\[ V_2 = \frac{9473.9 \times 10^4}{10.13 \times 10^4} = \frac{9473.9}{10.13} \]
\[ V_2 \approx 93.42 \, \text{cm}^3 \]

### Final Answer:
The volume at a pressure of \( 10.13 \times 10^4 \, \text{N/m}^2 \) is approximately \( 93.42 \, \text{cm}^3 \).

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