Question

A toy balloon has an internal pressure of 1.05 atm and a volume of 5.0 L. If the temperature where the balloon is released is 20º C, what will be the new volume when the balloon rises to an altitude where the pressure is 0.65 atm and the temperature is –15 ºC?

Answer 1

[AI-generated answer — Bot GPT 3.5]

First, we can use the ideal gas law equation to find the initial number of moles of gas in the balloon:

PV = nRT

Where:
P = pressure (1.05 atm)
V = volume (5.0 L)
n = number of moles of gas
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (20 ºC + 273 = 293 K)

Plugging in the values:

(1.05 atm)(5.0 L) = n(0.0821 L.atm/mol.K)(293 K)

n = (1.05*5) / (0.0821*293) ≈ 0.211 moles

Now, we can use the ideal gas law equation again to find the final volume at the new altitude and temperature:

PV = nRT

Where:
P = pressure (0.65 atm)
n = number of moles of gas (0.211 moles)
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (-15 ºC + 273 = 258 K)
V = final volume

Plugging in the values:

(0.65 atm)(V) = (0.211 moles)(0.0821 L.atm/mol.K)(258 K)

V = (0.211*0.0821*258) / 0.65 ≈ 6.55 L

Therefore, the new volume of the balloon when it rises to an altitude where the pressure is 0.65 atm and the temperature is -15 ºC will be approximately 6.55 L.