What is the compressibility factor (Z) for 0.02 mole of a van der Waals’ gas at pressure of 0.1 atm. Assume the size of gas molecules is negligible. Given: RT = 20 L atm mol-1 and a = 1000 atm L2 mol-2

Question

What is the compressibility factor (Z) for 0.02 mole of a van der Waals’ gas at pressure of 0.1 atm. Assume the size of gas molecules is negligible. Given: RT = 20 L atm mol-1 and a = 1000 atm L2 mol-2 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(0.1 + (1000×(0.02)2/V2))V = 20 ×0.02 0.1 V2 - 0.4 V + 0.4 = 0 V2 - 4V + 4 = 0 V = 2L Z = (PV/nRT) = (0.1×2/20×0.02) = 0.5

Q. What is the compressibility factor $(Z)$ for $0.02$ mole of a van der Waals’ gas at pressure of $0.1$ atm. Assume the size of gas molecules is negligible.
Given: $RT = 20 L$ atm mol $^{- 1}$ and $a = 1000$ atm $L^{2} m o l^{- 2}$

$(0.1 + \frac{1000 \times ( 0.02 ) ^{2}}{V ^{2}}) V = 20 \times 0.02$
$0.1 V^{2} - 0.4 V + 0.4 = 0$
$V^{2} - 4 V + 4 = 0$
$V = 2 L$
$Z = \frac{P V}{n RT}$
$= \frac{0.1 \times 2}{20 \times 0.02} = 0.5$

Q. What is the compressibility factor (Z) for 0.02 mole of a van der Waals’ gas at pressure of 0.1 atm. Assume the size of gas molecules is negligible. Given: RT=20L atm mol−1 and a=1000 atm L2mol−2

(0.1+V21000×(0.02)2​)V=20×0.02 0.1V2−0.4V+0.4=0 V2−4V+4=0 V=2L Z=nRTPV​ =20×0.020.1×2​=0.5

Q. What is the compressibility factor $(Z)$ for $0.02$ mole of a van der Waals’ gas at pressure of $0.1$ atm. Assume the size of gas molecules is negligible.
Given: $RT = 20 L$ atm mol $^{- 1}$ and $a = 1000$ atm $L^{2} m o l^{- 2}$

$(0.1 + \frac{1000 \times ( 0.02 ) ^{2}}{V ^{2}}) V = 20 \times 0.02$
$0.1 V^{2} - 0.4 V + 0.4 = 0$
$V^{2} - 4 V + 4 = 0$
$V = 2 L$
$Z = \frac{P V}{n RT}$
$= \frac{0.1 \times 2}{20 \times 0.02} = 0.5$