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Standard X

Physics

Ideal Gas Equation

When 2 g of a...

Question

When 2 g of a gas is introduced into an evacuated flask kept at $25^{\circ} C$ , the pressure is found to be 1 atmosphere. If 3 g of another gas B is then added to the same flask, the total pressure becomes 0.5 atmosphere. Assuming ideal gas behaviour, calculate the ratio of molecular mass $M_{A} : M_{B}$ .

3:1

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1:3

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Solution

The correct option is B
1:3

Applying ideal gas equation n moles,

PV= nRT,

$P V = \frac{g}{M} R T \Rightarrow M = \frac{g R T}{P V}$

Let the molecular mass of two gases be $M_{A} a n d M_{B}$

Therefore, $M_{A} = 2 \times \frac{R T}{1 \times V} a n d M_{B} = 3 \times \frac{R T}{0.5 \times V}$

$\Rightarrow \frac{M_{A}}{M_{B}} = \frac{2 R T}{V} \times \frac{0.5 V}{3 R T} = \frac{2 \times 0.5}{3} = \frac{1}{3}$

Ratio of molecular mass $M_{A} : M_{B} = 1 : 3$

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When 2 g of a gas is introduced into an evacuated flask kept at 25∘C, the pressure is found to be 1 atmosphere. If 3 g of another gas B is then added to the same flask, the total pressure becomes 0.5 atmosphere. Assuming ideal gas behaviour, calculate the ratio of molecular mass MA:MB.

The correct option is B 1:3 Applying ideal gas equation n moles, PV= nRT, PV=gMRT⇒M=gRTPV Let the molecular mass of two gases be MA and MB Therefore, MA=2×RT1×V and MB=3×RT0.5×V ⇒MAMB=2RTV×0.5V3RT=2×0.53=13 Ratio of molecular mass MA:MB=1:3