A certain quantity of real gas occupies a volume of 30.15 dm3 at 100 atm and 500 K when its compressibility factor is 1.07. Its volume at 300 atm and 300K (When its compressibility factor is 1.4) is _____ x 10-4 dm3 (Nearest integer)
Correct Answer: 392
Solution and Explanation
We can use the relation for the compressibility factor \( Z \) of a real gas: \[ Z = \frac{PV}{nRT}. \] Given the compressibility factor, we can modify the ideal gas equation: \[ V = \frac{ZnRT}{P}. \] For two different states of the gas, we can set up the following equation: \[ \frac{V_2}{V_1} = \frac{Z_2 P_1 T_1}{Z_1 P_2 T_2}. \] Given values:
- \( V_1 = 0.15 \, \text{dm}^3 \)
- \( P_1 = 100 \, \text{atm} \)
- \( T_1 = 500 \, \text{K} \)
- \( Z_1 = 1.07 \)
- \( P_2 = 300 \, \text{atm} \)
- \( T_2 = 300 \, \text{K} \)
- \( Z_2 = 1.4 \)
Substitute these values into the equation: \[ V_2 = V_1 \times \frac{Z_2 P_1 T_1}{Z_1 P_2 T_2}. \] \[ V_2 = 0.15 \times \frac{1.4 \times 100 \times 500}{1.07 \times 300 \times 300}. \] \[ V_2 \approx 0.15 \times \frac{70000}{96300} \approx 0.1089 \, \text{dm}^3. \] \[ V_2 \approx 108.9 \times 10^{-3} \, \text{dm}^3. \] Thus, the volume of the gas at 300 atm and 300 K is approximately \( 108.9 \times 10^{-3} \, \text{dm}^3 \)
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