High School

A balloon contains 38.2 g of [tex]$H_2$[/tex] gas at a volume of 482 L. What is the volume of the gas if 10.0 g is let out?

Answer :

We need to use the gas law equation PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is temperature.

First, we need to find the number of moles of gas in the balloon. We can do this using the formula:
n = m/M
where n is the number of moles, m is the mass of the gas (38.2g in this case), and M is the molar mass of the gas (2.016 g/mol for H2).
n = 38.2g / 2.016 g/mol = 18.95 mol
Now we can use the gas law equation to find the initial volume of the gas:
PV = nRT
V = nRT/P
where P is the pressure of the gas (which we don't know, but we can assume it's constant), T is the temperature (which

we also don't know, but we can assume it's constant), and R is the gas constant (0.0821 L·atm/mol·K).


V = (18.95 mol)(0.0821 L·atm/mol·K)(T/P)
Next, we need to find the volume of the gas when 10.0g is left out. To do this, we need to find the new number of moles of gas:
n' = (m - 10.0g)/M
n' = (38.2g - 10.0g) / 2.016 g/mol = 14.89 mol
Finally, we can use the gas law equation again to find the new volume of the gas:
V' = n'RT/P
V' = (14.89 mol)(0.0821 L·atm/mol·K)(T/P)
So the answer to the question is the difference between the initial volume and the new volume:
Volume of gas left out = V - V' = (18.95 mol)(0.0821 L·atm/mol·K)(T/P) - (14.89 mol)(0.0821 L·atm/mol·K)(T/P)

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