A balloon contains 10.0g of neon gas. With T kept constant, 10.0g of argon gas is added, and the volume of the balloon expands by less than 2 times. Why?
To understand why the volume of the balloon expands by less than 2 times when 10.0g of argon gas is added to the 10.0g of neon gas, we need to consider the properties of gases and the concept of molar mass.
First, let's understand the relationship between the amount of gas, its volume, and the conditions under which it is kept. This relationship is described by the ideal gas law, which states:
PV = nRT
where:
P represents the pressure of the gas,
V represents the volume of the gas,
n represents the number of moles of the gas,
R is the ideal gas constant, and
T is the temperature of the gas.
In this case, we are keeping the temperature (T) constant, so we can simplify the equation to:
PV = constant
Now, let's consider the molar mass of the gases involved. Molar mass is the mass of one mole of a substance, and it is expressed in grams per mole. The molar mass of neon (Ne) is approximately 20.18 g/mol, and the molar mass of argon (Ar) is approximately 39.95 g/mol.
When the 10.0g of neon is added to the balloon, we can calculate the number of moles of neon gas using its molar mass:
moles of neon (n) = mass of neon (m) / molar mass of neon (M)
= 10.0g / 20.18 g/mol
≈ 0.495 moles
Similarly, when the 10.0g of argon is added, we can calculate the number of moles of argon gas using its molar mass:
moles of argon (n) = mass of argon (m) / molar mass of argon (M)
= 10.0g / 39.95 g/mol
≈ 0.250 moles
Since the volume (V) is inversely proportional to the number of moles (n), according to the ideal gas law, the addition of argon gas with approximately half the number of moles compared to neon gas will result in a smaller increase in volume.
Therefore, the volume of the balloon expands by less than 2 times when 10.0g of argon gas is added because argon gas has a higher molar mass and fewer moles compared to neon gas.