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? \((R=0.082~L~atm~mol^{-1}K^{-1})\)

• A. \(1.15 \)
• B. \(0.95 \)
• C. \(1.10 \)
• D. \(0.91 \)

Hint:

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.

Answer 1

The Correct Option is C

Step 1: Know the compressibility factor.
The factor $Z$ tells how far a real gas drifts from ideal behaviour. It is defined as $Z=\frac{PV}{nRT}$. If $Z=1$ the gas is ideal.
Step 2: List the given data.
$P=2.71$ atm, $V=10$ L, $n=1$ mol, $T=300$ K, and $R=0.082$ L atm mol$^{-1}$K$^{-1}$.
Step 3: Work out the bottom of the fraction.
\[ nRT=1\times0.082\times300=24.6 \]
Step 4: Work out the top of the fraction.
\[ PV=2.71\times10=27.1 \]
Step 5: Divide to get Z.
\[ Z=\frac{27.1}{24.6}\approx1.10 \]
Step 6: Read the meaning.
Since $Z>1$ the gas is a bit harder to compress than an ideal gas. \[ \boxed{Z\approx1.10} \]