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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.
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