Final answer:
The internal pressure of a balloon with helium at -50.0°C, expanded to 20 times its original volume, would be 0.0394 atm. Considering the laboratory pressure remains at 1.00 atm, the gauge pressure would be -0.9606 atm, indicating the internal pressure is below atmospheric pressure, causing the balloon to contract.
Step-by-step explanation:
To answer these questions, we will apply the ideal gas law and concepts of pressure in gases. The ideal gas law is PV = nRT, where P is the pressure, V is the volume, T is the temperature (in Kelvin), n is the number of moles, and R is the ideal gas constant. First, we have to convert all temperatures to Kelvin by adding 273.15 to the Celsius values.
Part (a)
The pressure inside the balloon with helium at -50.0°C, when the volume is 20 times the original volume, can be determined using the combined gas law, which is P1V1/T1 = P2V2/T2. In this case, the initial pressure (P1) is 1.00 atmosphere, and the initial temperature (T1) is 10.0°C or 283.15 K. The final temperature (T2) is -50.0°C or 223.15 K.
Since the volume changes to 20 times the initial volume (V2 = 20V1), but the number of moles and the gas constant remain the same, we can solve for the final pressure (P2) inside the balloon:
P2 = P1V1T2 / (V2T1)
P2 = (1.00 atm × 1V1 × 223.15 K) / (20V1 × 283.15 K)
By canceling out V1, we find:
P2 = 223.15 / (20 × 283.15)
P2 = 0.0394 atm
Part (b)
The gauge pressure is the difference between the internal pressure of the balloon and the atmospheric pressure outside. Since the lab pressure remains at 1.00 atmosphere and the internal pressure of the balloon is 0.0394 atm, the gauge pressure is:
Gauge pressure = P2 - 1.00 atm
Gauge pressure = 0.0394 atm - 1.00 atm
Gauge pressure = -0.9606 atm
Note that a negative gauge pressure indicates that the internal pressure is lower than the external pressure. In practice, because the balloon is flexible, it would shrink until the pressures equalize.