Question

A weather balloon contains 13.0 L of helium at a pressure of 1.01 atm and a temperature of 21.0°C. If this had

Answer 1

Answer:

2.45atm

Explanation:

Use Combined Gas Law: $\frac{P_{1}V_{1}}{T_{1}} =\frac{P_{2}V_{2}}{T_{2}}$

Solve For P₂: $P_{2}=\frac{P_{1}V_{1}T_{2}}{T_{1}V_{2}}$

Use Kelvin: 21.0C = 294.15K 29.0 = 302.15K

(1.01atm)(13.0L)(302.15K)=3967.2295

(294.15K)(5.50L)=1617.825

3967.2295/1617.825=2.45219940352 = 2.45 atm