Question

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?

Answer 1

3.79L

Explanation:

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

• `PV=nRT`

Where for air in the lungs

• `"Pressure" (P)= 101.3kPa = 1atm`

• `"No. of moles"(n)=0.153`

• `"Temperature"(T)=302K`

• `"Universal gas constant " (R)= 0.082Latm K^-1mol^-1`

• `"Volume of air" (V) =?`

`:.V=(nRT)/P=(0.153xx0.082xx302)/1L=3.79L`