The pressure exerted by `6.0 g` of methane gas in a `0.03 m^(3)` vessel at `129^(@)C` is: (Atomic masses of `C= 12.01, H= 1.01 and R= 8.314 JK^(-1)mol^(-1)`)

Options

A. 13409 Pa
B. 41648 Pa
C. 31684 Pa
D. 215216 Pa

Answer

To solve the problem of finding the pressure exerted by 6.0 g of methane gas in a 0.03 m³ vessel at 129°C, we will use the Ideal Gas Law, which is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) = pressure (in Pascals)
- \( V \) = volume (in m³)
- \( n \) = number of moles of gas
- \( R \) = universal gas constant (8.314 J/(K·mol))
- \( T \) = temperature (in Kelvin)

### Step-by-Step Solution:

1. **Convert the mass of methane to moles:**
- The molecular formula for methane is \( CH_4 \).
- Calculate the molecular weight of methane:
\[ \text{Molecular weight of } CH_4 = 12.01 \, (\text{C}) + 4 \times 1.01 \, (\text{H}) = 12.01 + 4.04 = 16.05 \, \text{g/mol} \]
- Convert the mass of methane from grams to moles:
\[ n = \frac{\text{mass}}{\text{molecular weight}} = \frac{6.0 \, \text{g}}{16.05 \, \text{g/mol}} \approx 0.373 \, \text{mol} \]

2. **Convert the temperature from Celsius to Kelvin:**
- The temperature in Celsius is 129°C. To convert to Kelvin:
\[ T = 129 + 273 = 402 \, \text{K} \]

3. **Use the Ideal Gas Law to find the pressure:**
- Rearranging the Ideal Gas Law to solve for pressure \( P \):
\[ P = \frac{nRT}{V} \]
- Substitute the values into the equation:
\[ P = \frac{(0.373 \, \text{mol}) \times (8.314 \, \text{J/(K·mol)}) \times (402 \, \text{K})}{0.03 \, \text{m}^3} \]

4. **Calculate the pressure:**
- First, calculate the numerator:
\[ (0.373) \times (8.314) \times (402) \approx 1230.56 \, \text{J} \]
- Now divide by the volume:
\[ P \approx \frac{1230.56 \, \text{J}}{0.03 \, \text{m}^3} \approx 41018.67 \, \text{Pa} \]
- Rounding off, we get:
\[ P \approx 41019 \, \text{Pa} \text{ or } 41019 \, \text{Pascals} \]

### Final Answer:
The pressure exerted by 6.0 g of methane gas in a 0.03 m³ vessel at 129°C is approximately **41019 Pascals**.