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Ideal Gas Equation

The compressi...

Question

The compressibility factor (Z) of one mole of a van der Waals' gas of negligible 'a ' value is:

A

1

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B

\frac{b p}{R T}

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C

1 + \frac{b p}{R T}

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D

1 - \frac{b p}{R T}

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Solution

The correct option is C $1 + \frac{b p}{R T}$

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Similar questions

Q. The compressibility factor

(Z)

of one mole of a van der Waals' gas of negligible

^{'} a^{'}

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 :

R T = 20 L a t m m o l^{- 1}

a = 1000 a t m L^{2} m o l^{- 2}

Q. Assertion :The increase in compressibility factor with increasing pressure is due to

a

Z = 1 + \frac{b P}{R T}

for real gases can be obtained by neglecting

\frac{a}{V^{2}}

term in van der Waal's equation.

Q. Which of the following statements are correct?
I. Rise in compressibility factor

Z

with increase in pressure is due to '

a

'.
II. Rise in compressibility factor

Z

with increase in pressure is due to '

b

'.
III. Ideal gas do not exist but is a useful concept.
IV. For 1 mole of a van der Waal's gas,

Z = 1 + \frac{b P}{R T} - \frac{a}{R T V} + \frac{a b}{R T V^{2}}

Q. The compressibility factor for a real gas is expressed by,

z = 1 + \frac{B P}{R T}

. The value of B at 500 K and 600 bar is 0.0169 L/mol. Find the molar volume of the gas

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