Question 221 of 513
What number of moles of oxygen would exert a pressure of 10 atm at 320 K in an 8.2 dm\(^{2}\) cylinder?
[R = 0.082 atm dm\(^{3}\) mol\(^{-1}\) K\(^{-1}\)]
- A. 0.32
- B. 1.56
- C. 3.13
- D. 31.25
Correct Answer:
C
Explanation
To determine the number of moles of oxygen that would exert a pressure of 10 atm at 320 K in an 8.2 dm³ cylinder, we can use the Ideal Gas Law, which is expressed as:
\[ PV = nRT \]
Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in dm³)
- \( n \) = number of moles (in mol)
- \( R \) = ideal gas constant (in atm dm³ mol\(^{-1}\) K\(^{-1}\))
- \( T \) = temperature (in K)
### Step-by-Step Solution:
1. **Identify the given values:**
- Pressure, \( P = 10 \, \text{atm} \)
- Volume, \( V = 8.2 \, \text{dm}^3 \)
- Temperature, \( T = 320 \, \text{K} \)
- Ideal gas constant, \( R = 0.082 \, \text{atm dm}^3 \text{mol}^{-1} \text{K}^{-1} \)
2. **Rearrange the Ideal Gas Law to solve for \( n \):**
\[
n = \frac{PV}{RT}
\]
3. **Substitute the known values into the equation:**
\[
n = \frac{(10 \, \text{atm}) \times (8.2 \, \text{dm}^3)}{(0.082 \, \text{atm dm}^3 \text{mol}^{-1} \text{K}^{-1}) \times (320 \, \text{K})}
\]
4. **Calculate the denominator:**
\[
0.082 \, \text{atm dm}^3 \text{mol}^{-1} \text{K}^{-1} \times 320 \, \text{K} = 26.24 \, \text{atm dm}^3 \text{mol}^{-1}
\]
5. **Now calculate \( n \):**
\[
n = \frac{82 \, \text{atm dm}^3}{26.24 \, \text{atm dm}^3 \text{mol}^{-1}} \approx 3.12 \, \text{mol}
\]
6. **Round to two decimal places:**
\[
n \approx 3.13 \, \text{mol}
\]
### Conclusion:
The number of moles of oxygen that would exert a pressure of 10 atm at 320 K in an 8.2 dm³ cylinder is approximately **3.13 moles**. Therefore, the correct option is **C**.
### Explanation of Other Options:
- **Option A (0.32)**: This value is too low and does not reflect the conditions given. It suggests a very small amount of gas, which would not be sufficient to exert 10 atm in the specified volume.
- **Option B (1.56)**: This value is also too low. It does not account for the high pressure of 10 atm and the volume of 8.2 dm³, leading to an underestimation of the number of moles required.
- **Option D (31.25)**: This value is excessively high. It would imply an unrealistically large number of moles for the given pressure and volume, which contradicts the Ideal Gas Law.
### Revision Summary:
- Use the Ideal Gas Law \( PV = nRT \) to find the number of moles.
- Rearrange the formula to solve for \( n \): \( n = \frac{PV}{RT} \).
- Substitute known values carefully and perform calculations step-by-step.
- Ensure units are consistent and check calculations for accuracy.