A 4.0 L vessel contains He at 300 K and 2.0 atm. 0.10 mol of N2 is added at the same temperature. Taking R = 0.0821 L atm K^-1 mol^-1, the final total pressure is nearest to

JEE Advanced levelSingle correct optionDifficulty 3 of 3ig201

A $4.0 L$ vessel contains $He$ at $300 K$ and $2.0 atm$ . $0.10 mol$ of $N X_{2}$ is added at the same temperature. Taking $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , the final total pressure is nearest to

p_{N X_{2}} = \frac{0.10 \times 0.0821 \times 300}{4.0} = 0.616 atm

. By Dalton's law

P = 2.0 + 0.616 = 2.62 \approx 2.6 atm