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?

1 Answer
May 13, 2016

Answer:

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#