Question

What is the volume of $2.3$ moles of an ideal gas at $300K$ and a pressure of $1.1$ atm?

• A. $\frac{\left(2.3\right)\left(0.0820\right)\left(300\right)}{\left(1.1\right)}$ L
• B. $\frac{\left(1.1\right)\left(2.3\right)\left(0.0820\right)}{\left(300\right)}$ L
• C. $\frac{\left(2.3\right)\left(0.0820\right)}{\left(300\right)\left(1.1\right)}$ L
• D. $\frac{\left(300\right)\left(0.0820\right)}{\left(2.3\right)\left(1.1\right)}$ L
• E. $\frac{\left(2.3\right)\left(1.1\right)\left(300\right)}{\left(0.0820\right)}$ L

Answer 1

Answer: (A)

$\frac{\left(2.3\right)\left(0.0820\right)\left(300\right)}{\left(1.1\right)}$ L

Given that,

Number of moles, $n=2.3mol$

Temperature, $T=300K$

Pressure, $P=1.1atm$

From Ideal gas equation,

$Volume,V=\frac{nRT}{P}$

$=\frac{2.3\times 0.0820\times 300}{1.1}L$

where,

$R=0.082Latmmo{l}^{-1}{K}^{-1}$

Hence, option $\left(A\right)$ is the correct answer.