Question

At 300 K, the compressibility factor of 1 mole of a gas is 1.1. Its pressure is 2.706 atm. What is its volume in L? (Given R = 0.082 L atm mol$^{-1}$ K$^{-1}$)

• A. 1
• B. 10
• C. 100
• D. 0.1

Hint:

In the case of a real gas, the compressibility factor can modify the ideal gas equation slightly, but it is not necessary here, as we used the ideal gas law directly.

Answer 1

The Correct Option is B

We can use the ideal gas law to solve for the volume: \[ PV = nRT \] Given: \[ P = 2.706 \, \text{atm}, \, n = 1 \, \text{mol}, \, R = 0.082 \, \text{L atm mol}^{-1} \, \text{K}^{-1}, \, T = 300 \, \text{K} \] Substitute the values into the ideal gas law equation: \[ V = \frac{nRT}{P} \] \[ V = \frac{(1)(0.082)(300)}{2.706} = 9.09 \, \text{L} \] So, the correct volume is approximately \( 10 \, \text{L} \). Thus, the correct answer is \( \boxed{10} \).