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?

Cause of the ratio

The volume of a gas is directly proportional to the number of moles of the gas, assuming constant temperature and pressure. Therefore, in order for the volume of the balloon to expand by less than 2 times, the number of moles of gas in the balloon must increase less than 2 times.

To compare the number of moles of neon gas and argon gas, we need to use their molar masses. The molar mass of neon (Ne) is approximately 20.2 g/mol, and the molar mass of argon (Ar) is approximately 39.9 g/mol.

First, let's calculate the number of moles of neon gas:

Moles of neon gas = mass of neon gas / molar mass of neon gas
Moles of neon gas = 10.0 g / 20.2 g/mol
Moles of neon gas ≈ 0.495 moles

Now, let's calculate the number of moles of argon gas that is added:

Moles of argon gas = mass of argon gas / molar mass of argon gas
Moles of argon gas = 10.0 g / 39.9 g/mol
Moles of argon gas ≈ 0.251 moles

The total number of moles of gas in the balloon after adding the argon gas is approximately:

Total moles of gas = moles of neon gas + moles of argon gas
Total moles of gas ≈ 0.495 moles + 0.251 moles
Total moles of gas ≈ 0.746 moles

Since the volume of the balloon expands by less than 2 times, we can deduce that the volume of the balloon increased by a factor less than 2, while the number of moles of gas increased by approximately 0.746 moles. This smaller increase in moles of gas compared to the increase in volume implies that the argon gas particles are larger, occupy more space per mole, making the volume increase less than doubling.

To understand why the volume of the balloon expands by less than 2 times when 10.0g of argon gas is added to the balloon containing 10.0g of neon gas, we need to consider the ideal gas law.

The ideal gas law is given by the equation:

PV = nRT

where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of gas
R is the ideal gas constant
T is the temperature of the gas

Since the temperature (T) is kept constant in this scenario, we can simplify the ideal gas law to:

PV = constant

Dividing this equation by the number of moles (n) on both sides, we get:

PV/n = constant

For a given amount of gas, the product of pressure (P) and volume (V) divided by the number of moles (n) is a constant.

Let's compare the initial state of the balloon with the final state after argon gas is added:

Initial state:
- Neon gas: 10.0g
- Volume: V1

Final state:
- Neon gas: 10.0g
- Argon gas: 10.0g
- Volume: V2 (expanded by less than 2 times)

Since the amount of neon gas remains the same, the number of moles of neon gas (n1) stays constant. However, when the same mass of argon gas is added, the number of moles of argon gas (n2) is greater than n1, as the molar mass of argon is greater than that of neon.

Therefore, for the final state, we have:

(P*V1)/n1 = (P*V2)/n2

We can see that n2 is greater than n1, so for the equation to hold true, V2 must be less than V1.

This means that when the same mass of argon gas is added to the balloon, the volume increases by less than 2 times because the added argon gas occupies less volume compared to the neon gas.

In summary, the difference in volume expansion can be attributed to the fact that the molar mass of argon gas is greater than that of neon gas.