For a real gas at 25°C temperature and high pressure (99 bar) the value of the compressibility factor is 2, so the value of van der Waal’s constant ‘b’ should be ______ × 10–2 L mol–1 (Nearest integer).
(Given R=0.083 L bar K-1 mol-1)
(Given R=0.083 L bar K-1 mol-1)
Correct Answer: 25
Solution and Explanation
The correct answer is: 25.
For 1 mole at high pressure
P(V-b) = RT
PV-Pb = RT
\(\frac{PV}{RT}\)= 1+\(\frac{Pb}{RT}\)
Z = 1+\(\frac{Pb}{RT}\)
1 = \(\frac{99(b)}{0.083×298}\)
b = \(\frac{0.083 ×298}{99}\)≈0.249≈25×10-2
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Concepts Used:
Ideal Gas Equation
An ideal gas is a theoretical gas composed of a set of randomly-moving point particles that interact only through elastic collisions.
What is Ideal Gas Law?
The ideal gas law states that the product of the pressure and the volume of one gram molecule of an ideal gas is equal to the product of the absolute temperature of the gas and the universal gas constant.
PV=nRT
where,
P is the pressure
V is the volume
n is the amount of substance
R is the ideal gas constant
Ideal Gas Law Units
When we use the gas constant R = 8.31 J/K.mol, then we have to plug in the pressure P in the units of pascals Pa, volume in the units of m3 and the temperature T in the units of kelvin K.