Question:
At 300 K, one mole of a gas present in a 10 L flask exerted a pressure of 2.71 atm. What is its compressibility factor? \((R=0.082~L~atm~mol^{-1}K^{-1})\)
At 300 K, one mole of a gas present in a 10 L flask exerted a pressure of 2.71 atm. What is its compressibility factor? \((R=0.082~L~atm~mol^{-1}K^{-1})\)
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For an ideal gas, \(Z = 1\). If \(Z > 1\), the gas shows positive deviation from ideal behavior due to repulsive forces; if \(Z < 1\), it shows negative deviation due to attractive forces.
Updated On: Jun 7, 2026
- \(1.15 \)
- \(0.95 \)
- \(1.10 \)
- \(0.91 \)
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The Correct Option is C
Solution and Explanation
Concept:
The compressibility factor (\(Z\)) is a dimensionless quantity used in thermodynamics to describe the deviation of a real gas from ideal gas behavior. It is defined as the ratio of the actual molar volume of a gas to the molar volume of an ideal gas at the same temperature and pressure:
\[
Z = \frac{PV}{nRT}
\]
where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the universal gas constant, and \(T\) is the absolute temperature.
Step 1: Identify the given physical parameters from the problem statement.
* Pressure (\(P\)) = \(2.71\) atm * Volume (\(V\)) = \(10\) L * Moles (\(n\)) = \(1\) mole * Temperature (\(T\)) = \(300\) K * Gas Constant (\(R\)) = \(0.082\) L atm mol\(^{-1}\) K\(^{-1}\)
Step 2: Calculate the ideal volume (\(V_{ideal}\)) using the ideal gas law \(PV = nRT\).
The volume that one mole of an ideal gas would occupy under these conditions is: \[ V_{ideal} = \frac{nRT}{P} = \frac{1 \times 0.082 \times 300}{2.71} = \frac{24.6}{2.71} \approx 9.077 \text{ L} \]
Step 3: Calculate the compressibility factor (\(Z\)).
Substitute the actual values into the \(Z\) formula: \[ Z = \frac{PV}{nRT} = \frac{2.71 \times 10}{1 \times 0.082 \times 300} = \frac{27.1}{24.6} \approx 1.10 \]
Step 1: Identify the given physical parameters from the problem statement.
* Pressure (\(P\)) = \(2.71\) atm * Volume (\(V\)) = \(10\) L * Moles (\(n\)) = \(1\) mole * Temperature (\(T\)) = \(300\) K * Gas Constant (\(R\)) = \(0.082\) L atm mol\(^{-1}\) K\(^{-1}\)
Step 2: Calculate the ideal volume (\(V_{ideal}\)) using the ideal gas law \(PV = nRT\).
The volume that one mole of an ideal gas would occupy under these conditions is: \[ V_{ideal} = \frac{nRT}{P} = \frac{1 \times 0.082 \times 300}{2.71} = \frac{24.6}{2.71} \approx 9.077 \text{ L} \]
Step 3: Calculate the compressibility factor (\(Z\)).
Substitute the actual values into the \(Z\) formula: \[ Z = \frac{PV}{nRT} = \frac{2.71 \times 10}{1 \times 0.082 \times 300} = \frac{27.1}{24.6} \approx 1.10 \]
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