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Standard XII

Physics

Real Gas Equation

Pressure exer...

Question

Pressure exerted by $1 mole$ of methane in $0.25 litre$ container at $300 K$ using Van der Waal's equations?
Given $a = 2.25 {atm - L}^{2} / {mol}^{2}$
$b = 0.05 L/mol$

86 atm

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87.15 atm

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100 atm

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50 atm

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Solution

The correct option is B $87.15 atm$
We know,
For real gas,
$(P + \frac{n^{2} a}{V^{2}}) (V - n b) = n R T$
where,
$P =$ Pressure
$n =$ no of moles
$V =$ Volume
$(P + \frac{1^{2} \times 2.25}{0.25 \times 0.25}) (0.25 - 0.05) = 1 \times 0.0821 \times 300$
$\Rightarrow P = 87.15 atm$

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Real Gas Equation

Standard XII Physics

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Pressure exerted by 1 mole of methane in 0.25 litre container at 300 K using Van der Waal's equations? Given a=2.25 atm - L2/mol2 b=0.05 L/mol

The correct option is B 87.15 atmWe know, For real gas, (P+n2aV2)(V−nb)=nRT where, P= Pressure n= no of moles V= Volume (P+12×2.250.25×0.25)(0.25−0.05)=1×0.0821×300 ⇒P=87.15 atm

Pressure exerted by $1 mole$ of methane in $0.25 litre$ container at $300 K$ using Van der Waal's equations?
Given $a = 2.25 {atm - L}^{2} / {mol}^{2}$
$b = 0.05 L/mol$

The correct option is B $87.15 atm$
We know,
For real gas,
$(P + \frac{n^{2} a}{V^{2}}) (V - n b) = n R T$
where,
$P =$ Pressure
$n =$ no of moles
$V =$ Volume
$(P + \frac{1^{2} \times 2.25}{0.25 \times 0.25}) (0.25 - 0.05) = 1 \times 0.0821 \times 300$
$\Rightarrow P = 87.15 atm$