Question

A steel container with a movable piston contains 2.00 g of helium which was held at a constant temperature of 25 °C. Additional helium was pumped into the container and the piston adjusted so that the gas pressure remained constant. How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 3.00 L?A) 0.700 gB) 2.00 gC) 1.8 gD) 1.00 gE) 9.7 gF) 5.63 gG) 4.63 g

Answer 1

Answer: D) 1.00 g

Explanation:

According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

$V\propto n$

or,

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

where,

$V_1$ = initial volume of gas  = 2.00 L

$V_2$ = final volume of gas = 3.00 L

$n_1$ = initial moles of gas  =$\frac{\text {Given mass of helium}}{\text {molar mass of helium}}=\frac{2.00g}{4g/mol}=0.500mol$

$n_2$ = final moles of gas  = ?

Now we put all the given values in this formula, we get

$\frac{2.00L}{3.00L}=\frac{0.500mol}{n_2}$

$n_2=0.75mole$

Mass of helium =$moles\times {\text {molar mass}}=0.75\times 4=3.00g$

Thus mass of helium added = (3.00-2.00) g = 1.00 g