Question

A 4.00g sample of helium has a volume of 24.4L at a temperature of 25.0oC and a pressure of 1.00 atm. The volume of the helium is reduced to 10.4L, but the temperature and pressure of the gas are kept con- stant. What is the new quantity of the gas, in moles? Show all work using units and sig figs.

Answer 1

Answer:

0.41 moles.

Explanation:

Given that:

Mass of helium = 4.00 g

Initial Volume = 24.4 L

initial Temperature = 25.0 °C =( 25 + 273) = 298 K

initial Pressure = 1.00 atm

The volume was reduced to :

i.e

final volume of the helium - 10.4 L

Change in ΔV = 24.4 - 10.4 = 10.0 L

Temperature and pressure remains constant.

The new quantity of gas can be calculated by using the ideal gas equation.

PV = nRT

n = $\frac{PV}{RT}$

n = $\frac{1.00*10.0}{0.082057*298}$

n = 0.4089 moles

n = 0.41 moles.