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

Answer:

2.83 g

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:

$\frac{2.12}{0.120}=\frac{3.12}{n_2}$

$2.12n_2=0.12\cdot \:3.12$

$2.12n_2=0.3744$

$n_2=\frac{0.3744}{2.12}$

$n_2=0.17660$

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.