# The air in lungs has 0.153 mol of gas particles at 302 K and 101.3 kPa of pressure. What is the volume of the air?

May 13, 2016

3.79L

#### Explanation:

This can be solved applying equation of state for ideal gas, which i as follows

• $P V = n R T$

Where for air in the lungs

• $\text{Pressure} \left(P\right) = 101.3 k P a = 1 a t m$

• $\text{No. of moles} \left(n\right) = 0.153$

• $\text{Temperature} \left(T\right) = 302 K$

• $\text{Universal gas constant } \left(R\right) = 0.082 L a t m {K}^{-} 1 m o {l}^{-} 1$

• "Volume of air" (V) =?

$\therefore V = \frac{n R T}{P} = \frac{0.153 \times 0.082 \times 302}{1} L = 3.79 L$