# What is the volume of the container needed to store 0.8 moles of argon gas at 5.3 atm and 227°C?

Apr 30, 2016

$V = \frac{n R T}{P}$

#### Explanation:

$V = \frac{0.8 \cdot m o l \times 0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 m o {l}^{- 1} \cdot 500 \cdot K}{5.3 \cdot a t m}$

$=$ ??L

Apr 30, 2016

$= 6.19 L$

#### Explanation:

By Ideal gas equation we know,

$P V = n R T$

Where

• Pressure $P = 5.3 a t m$
• Volume V=?L
• Universal gas constant $R = 0.082 L a t m {K}^{-} 1 m o {l}^{-} 1$
• Number moles of gas $n = 0.8 m o l$
• The temperature of the gas $T = 227 + 273 = 500 K$

Putting inthe gas equation we get

$5.3 V = 0.8 \times 0.082 \times 500$

$V = \frac{0.8 \times 0.082 \times 500}{5.3} L = 6.19 L$