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Chemistry - Ideal Gas Question with Solution | TestHub

ChemistryIdeal GasIdeal Gas Equation & problemsMedium2 minQB

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Passage / Comprehension

Paragraph for Questions 10 and 11

A 4.92 L Duma's bulb contains nitrogen gas at 1.0 atm. On adding 0.05 mole of $A r$ gas it is necessary to cool the bulb to a temperature of 27° $C$ to maintain the same pressure. ( $R$ = 0.082 L atm/K mol)

The total no. of moles of $N_{2}$ in the Duma's bulb is

Options:

Answer:

Solution:

Let $n$ = moles of $N_{2}$ and $T_{1}$ the initial temperature. At $P$ = 1.0 atm and $V$ = 4.92 L: $P V$ = $n R T_{1}$ , so $n T_{1}$ = 4.92/0.082 = 60.

After adding 0.05 mol $A r$ and cooling to 27° $C$ (300 K) at the same $P$ and $V$ : ( $n$ + 0.05) x 300 = 60, so $n$ + 0.05 = 0.20.

$n$ = 0.15 mol.

Answer: (A).

Stream:JEESubject:ChemistryTopic:Ideal GasSubtopic:Ideal Gas Equation & problems

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