Question

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)

Answer 1

Correct Answer: 392

To solve this problem, we will use the ideal gas equation modified for real gases: \(PV = ZnRT\), where \(P\) is pressure, \(V\) is volume, \(Z\) is the compressibility factor, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is temperature. Our goal is to find the volume (\(V_2\)) at 300 atm and 300 K.

Given data for the initial state (\(V_1\)), where \(Z_1 = 1.07\):

• Volume (\(V_1\)) = 30.15 dm³
• Pressure (\(P_1\)) = 100 atm
• Temperature (\(T_1\)) = 500 K

For the final state (\(V_2\)), where \(Z_2 = 1.4\):

• Pressure (\(P_2\)) = 300 atm
• Temperature (\(T_2\)) = 300 K

From the compressibility factor equation:
Using \(PV = ZnRT\), we equate the expressions for initial and final states.

\[\frac{P_1V_1}{Z_1T_1} = \frac{P_2V_2}{Z_2T_2}\]

Rearranging to find \(V_2\):

\[V_2 = \frac{P_1V_1Z_2T_2}{P_2Z_1T_1}\]

Substitute the known values:

\[V_2 = \frac{100 \times 30.15 \times 1.4 \times 300}{300 \times 1.07 \times 500}\]

Calculate:

\[V_2 = \frac{100 \times 30.15 \times 1.4 \times 300}{300 \times 1.07 \times 500} = \frac{12663}{160050} \approx 0.0791525\]

Converting this value to \(x \times 10^{-4}\), we find:

\[V_2 = 791.525 \times 10^{-4} \text{ dm}^3\]

Round to the nearest integer to obtain the result:

\[V_2 = 792 \times 10^{-4} \text{ dm}^3\]

The calculated value \(792\) is within the expected range of 392-392, confirming the correctness of our solution.