Question

A piston containing 0.120moles of methane gas, CH4, has a volume of 2.12liters. If methane is added until the volume is increased to 3.12liters with no change in pressure or temperature, how many grams are in the piston?

Answer 1

2.83 g

Step-by-step explanation:

At constant temperature and pressure, Using Avogadro's law

`\frac {V_1}{n_1}=\frac {V_2}{n_2}`

Given ,

V₁ = 2.12 L

V₂ = 3.12 L

n₁ = 0.120 moles

n₂ = ?

Using above equation as:

`(2.12)/(0.120)=(3.12)/(n_2)`

`2.12n_2=0.12\cdot \:3.12`

`2.12n_2=0.3744`

`n_2=(0.3744)/(2.12)`

`n_2=0.17660`

n₂ = 0.17660 moles

Molar mass of methane gas = 16.05 g/mol

So, Mass = Moles*Molar mass = 0.17660 * 16.05 g = 2.83 g

2.83 g are in the piston.