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Standard XII

Chemistry

Dalton's Law of Partial Pressures

2 moles of an...

Question

$2$ moles of an ideal gas are contained in a $20$ litre vessel. If the temperature of gas is $300 K$ , then the pressure of the gas will be

$(1 atm = 10^{5} {N/m}^{2})$

1.2 {N/m}^{2}

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2.5 atm

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6.1 atm

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2.1 \times 10^{3} {N/m}^{2}

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Solution

The correct option is B $2.5 atm$
As per ideal gas equation,

$P V = n R T$

$P = \frac{n R T}{V} = \frac{2 \times 8.314 \times 300}{20 \times 10^{- 3}} = 2.5 atm$

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2 moles of an ideal gas are contained in a 20 litre vessel. If the temperature of gas is 300 K, then the pressure of the gas will be (1 atm=105 N/m2)

The correct option is B 2.5 atmAs per ideal gas equation, PV=nRT P=nRTV=2×8.314×30020×10−3=2.5 atm

$2$ moles of an ideal gas are contained in a $20$ litre vessel. If the temperature of gas is $300 K$ , then the pressure of the gas will be

$(1 atm = 10^{5} {N/m}^{2})$

The correct option is B $2.5 atm$
As per ideal gas equation,

$P V = n R T$

$P = \frac{n R T}{V} = \frac{2 \times 8.314 \times 300}{20 \times 10^{- 3}} = 2.5 atm$