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JEE Main levelNumerical answerDifficulty 2 of 3ig193

The amount of an ideal gas, in millimoles (one decimal), contained in a $2.0 L$ vessel at $400 K$ and $1.2 atm$ is (take $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ )

Numerical answer type, answer in mmol, tolerance 0.1.

Answer: 73.1 mmol

n = \frac{P V}{RT} = \frac{1.2 \times 2.0}{0.0821 \times 400} = 0.07308 mol = 73.1 mmol

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