Question

Under what conditions will a pure sample of an ideal gas not only exhibit a pressure of 1 atm, but also a concentration of 1 mole $litre^{- 1}$ ?(R = 0.082 litre atm $mol^{- 1}deg^{- 1}$ )

• A. At STP
• B. When V = 22.4
• C. When T = 12 K
• D. Impossible under any condition

Answer 1

Answer: (C)

The gas has a concentration of $\text{1} \, \text{mol/L}$ .
This means that $\text{V} \, \text{=} \, \text{1}$
Number of mol of gas $\text{(n)} \, \text{=} \, \text{1} \text{​} \, \text{mol}$
$\text{P} \, \text{=} \, \text{1} \, \text{atm}$
$\text{R} \, \text{=} \, \text{0} \text{.082} \, \text{L} \, \text{atm} \, \text{mol} \, \text{K}^{- \text{1}}$
$\text{T} \, \text{=} \, \text{?}$
So $PV=nRT$
So $T=\frac{P V}{n R}$
$\text{=} \, \frac{\text{1} \, \text{atm} \, \text{\times } \, \text{1} \, \text{L}}{\text{1} \, \text{mol} \, \text{\times } \, \text{0} \text{.082} \, \text{L} \, \text{atm} \, \text{mol}^{- \text{1}} \, \text{K}^{- \text{1}}}$
$\text{=} \, \frac{\text{1}}{\text{0} \text{.082}} \, \text{=} \, \text{12} \, \text{K}$