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Standard XII

Chemistry

Liquefaction of Gases and Critical Constants

The van der W...

Question

The van der Waal's constants for a gas are $a = 3.6 a t m L^{2} {m o l}^{2}, b = 0.6 L {m o l}^{- 1}$ . If $R = 0.08 L a t m K^{- 1} {m o l}^{- 1}$ and the Boyle's temperature ( $K$ ) is $T_{0}$ of this gas, then what is the value of $\frac{T_{B}}{15}$ ?

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Solution

$\begin{matrix} a = 3.6 a t m L^{2} m o l^{2} & b = 0.6 L m o l^{- 1} T_{B} = \frac{a}{R b} \Rightarrow \frac{3.6}{6.08 \times 0.6} \frac{a t m L^{2} m o l^{- 2}}{L m o l^{- 1} L a t m K^{- 1} m o l^{- 1}} T_{B} = 75 K & \frac{T_{B}}{15} = \frac{75}{15} = 5 \end{matrix}$

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Similar questions

Q. The van der Waals’ constants for a gas are;

a = 3.6 {atm L}^{2} {mol}^{2}

b = 0.6 {L mol}^{- 1}

. If

R = 0.08 {L atm K}^{- 1}

. If the Boyle's temperature

(K)

of this gas is

T_{B}

, then what is the value of

\frac{T_{B}}{15}

Q. The van der Waals’ constants for a gas are;

a = 3.6 {atm L}^{2} {mol}^{2}

b = 0.6 {L mol}^{- 1}

. If

R = 0.08 {L atm K}^{- 1}

. If the Boyle's temperature

(K)

of this gas is

T_{B}

, then what is the value of

\frac{T_{B}}{15}

For one mole of a van der Waal's gas when

b = 0

and

T = 300 K

, the

P V

1 / V

plot is shown above.

The value of the van der Waal's constant

a

(in atm

l i t r e^{2} m o l^{- 2})

is :

Q. For one mole of a van der Waal's gas when b = 0 and T = 300 K, the PV vs

\frac{1}{V}

plot is shown below. The value of the van der Waal's constant a

({atm. liter}^{2} {mol}^{- 2})

is:

Calculate the pressure of one mole of nitrogen gas in a 10 L cylinder at $50^{\circ} C$ using the Van der Waal's equation.Also find the value of ideal gas pressure.(Van der Waal's constants a and b for nitrogen are $1.4 L^{2} a t m m o l^{- 2} a n d 0.04 L^{2} m o l^{- 1}$ respectively) .

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The van der Waal's constants for a gas are a=3.6atmL2mol2,b=0.6Lmol−1. If R=0.08LatmK−1mol−1 and the Boyle's temperature (K) is T0 of this gas, then what is the value of TB15?

a=3.6atmL2mol2b=0.6Lmol−1TB=aRb⇒3.66.08×0.6atmL2mol−2Lmol−1LatmK−1mol−1TB=75KTB15=7515=5

The van der Waal's constants for a gas are $a = 3.6 a t m L^{2} {m o l}^{2}, b = 0.6 L {m o l}^{- 1}$ . If $R = 0.08 L a t m K^{- 1} {m o l}^{- 1}$ and the Boyle's temperature ( $K$ ) is $T_{0}$ of this gas, then what is the value of $\frac{T_{B}}{15}$ ?

$\begin{matrix} a = 3.6 a t m L^{2} m o l^{2} & b = 0.6 L m o l^{- 1} T_{B} = \frac{a}{R b} \Rightarrow \frac{3.6}{6.08 \times 0.6} \frac{a t m L^{2} m o l^{- 2}}{L m o l^{- 1} L a t m K^{- 1} m o l^{- 1}} T_{B} = 75 K & \frac{T_{B}}{15} = \frac{75}{15} = 5 \end{matrix}$