Question:
Calculate the compressibility factor of \(1\) mole of a certain real gas if it occupies \(0.4\text{ dm}^3\) at \(300\) K and \(40\) atm. [R \(= 0.082\) atm \(\text{dm}^3\text{K}^{-1}\text{mol}^{-1}\)]
Calculate the compressibility factor of \(1\) mole of a certain real gas if it occupies \(0.4\text{ dm}^3\) at \(300\) K and \(40\) atm. [R \(= 0.082\) atm \(\text{dm}^3\text{K}^{-1}\text{mol}^{-1}\)]
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Compressibility factor is Z = PV / nRT, which compares the real molar volume with the ideal one.
Updated On: Oct 1, 2026
- \(0.45\)
- \(0.65\)
- \(0.85\)
- \(1.00\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
For an ideal gas, \(Z = 1\). The deviation of \(Z\) from 1 shows how far a real gas departs from ideal behaviour.
Step 2: Key Formula or Approach:
\[ Z = \frac{PV}{nRT} \]
Step 3: Calculate.
\(P = 40\) atm, \(V = 0.4\) dm\(^3\), \(n = 1\), \(T = 300\) K, \(R = 0.082\).
\[ PV = 40\times 0.4 = 16 \]
\[ nRT = 0.082\times 300 = 24.6 \]
\[ Z = \frac{16}{24.6} = 0.65 \]
Step 4: Check the options.
\(Z = 1\) would be an ideal gas. Here \(Z < 1\), so attractive forces dominate.
Final Answer:
The compressibility factor is 0.65, option (B). \[ \boxed{Z = 0.65} \]
For an ideal gas, \(Z = 1\). The deviation of \(Z\) from 1 shows how far a real gas departs from ideal behaviour.
Step 2: Key Formula or Approach:
\[ Z = \frac{PV}{nRT} \]
Step 3: Calculate.
\(P = 40\) atm, \(V = 0.4\) dm\(^3\), \(n = 1\), \(T = 300\) K, \(R = 0.082\).
\[ PV = 40\times 0.4 = 16 \]
\[ nRT = 0.082\times 300 = 24.6 \]
\[ Z = \frac{16}{24.6} = 0.65 \]
Step 4: Check the options.
\(Z = 1\) would be an ideal gas. Here \(Z < 1\), so attractive forces dominate.
Final Answer:
The compressibility factor is 0.65, option (B). \[ \boxed{Z = 0.65} \]
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