At an altitude for which ambient pressure is #0.700*atm#, and a temperature of #270.15*K#, a balloon contains #3.00*mol# of gas... What is the volume of the balloon....?

Question

At an altitude for which ambient pressure is #0.700*atm#, and a temperature of #270.15*K#, a balloon contains #3.00*mol# of gas...  What is the volume of the balloon....?

Naomi Aronson · Accepted Answer

There is no question here, but your neighbour, the balloonist, has filled a balloon with approx. #100*L# volume.

Of course, he may have heated the contents of the balloon somehow, but we must assume #T=270*K#, so.......... #V=(nRT)/P=(3.00*molxx0.0821*(L*atm)/(K*mol)xx270.15*K)/(0.700*atm)# #~=100*L.# We have assumed the temperature, and know that the balloon expanded against a #0.700*atm# pressure. Because the balloon is apparently floating, what is the probable filling gas?

Luna Chen · Answer

undefined The volume of the balloon can be calculated using the ideal gas law equation:

$$PV = nRT$$

Where:
- $P$ is the pressure of the gas (in atm)
- $V$ is the volume of the gas (in liters)
- $n$ is the number of moles of gas
- $R$ is the ideal gas constant (0.0821 L·atm/(K·mol))
- $T$ is the temperature of the gas (in Kelvin)

Given:
- Pressure ($P$) = 0.700 atm
- Temperature ($T$) = 270.15 K
- Number of moles ($n$) = 3.00 mol

We can rearrange the ideal gas law to solve for volume ($V$):

$$V = \frac{{nRT}}{{P}}$$

$$V = \frac{{(3.00\ mol)(0.0821\ L·atm/(K·mol))(270.15\ K)}}{{0.700\ atm}}$$

$$V = \frac{{(3.00\ mol)(22.359\ L/mol)}}{{0.700}}$$

$$V = \frac{{67.077\ L}}{{0.700}}$$

$$V ≈ 95.82\ L$$

Therefore, the volume of the balloon is approximately 95.82 liters.