Question

A weather balloon containing helium with a volume of 410.0 L rises in the atmosphere and is cooled from 17∘

∘
C to -27∘
∘
C. The pressure on the gas is reduced from 110.0 kPa to 25.0 kPa. What is the volume of the gas at the lower temperature and pressure? What law did you use to solve?

Answer 1

The volume of the gas at the lower temperature and pressure is 457.72 L. The combined gas law was used to solve this problem.

Step-by-step explanation:

In order to solve this problem, we can use the combined gas law. The combined gas law allows us to calculate the effect of varying temperature and pressure on the volume of a gas sample, assuming the quantity of gas remains constant. The combined gas law equation is:

P₁V₁/T₁ = P₂V₂/T₂

where P₁ and T₁ are the initial pressure and temperature, P₂ and T₂ are the final pressure and temperature, and V₁ and V₂ are the initial and final volumes. Rearranging the equation, we get:

V₂ = (P₁V₁T₂)/(P₂T₁)

Using the given values, the initial pressure (P₁) is 110.0 kPa, the final pressure (P₂) is 25.0 kPa, the initial temperature (T₁) is 17°C converted to Kelvin (17 + 273.15), and the final temperature (T₂) is -27°C converted to Kelvin (-27 + 273.15). The initial volume (V₁) is 410.0 L. Substituting these values into the equation, we can solve for V₂ to find the volume of the gas at the lower temperature and pressure.

V₂ = (110.0 kPa * 410.0 L * (-27 + 273.15 K))/(25.0 kPa * (17 + 273.15 K)) = 457.72 L

Therefore, the volume of the gas at the lower temperature and pressure is 457.72 L.

Answer 2

V₂ = 1500 Liters ( 2 sig. figs.)

Step-by-step explanation:

Given the following gas law variables:

P₁ = 110.0KPa P₂ = 25KPa

V₁ = 410 Liters V₂ = ?

T₁ = 17°C ( = 290K) T₂ = -27°C ( = 248K)

P₁V₁/₁T₁ = P₂V₂/T₂ => V₂ = V₁(P₁/P₂)(T₂/T₂)

V₂ = 410L(110.0KPa/25KPa)(248k/290K) = 1542 L (calc. ans.)

V₂ = 1500 Liters ( 2 sig. figs.)