Question

A 250 mL flexible container begins at room temperature (25 °C) at a pressure of 30.07 inHg. The container is taken to a new altitude where the pressure is 9.96 inHg at a temperature of 3.7 °C. What would the volume of the container be in these new conditions?

Answer 1

Therefore, the volume of the container in the new conditions would be approximately 221.96 mL.

Step-by-step explanation:

To determine the volume of the container in the new conditions, we can use the ideal gas law, which states:

PV = nRT

Where:

P = pressure

V = volume

n = number of moles of gas

R = ideal gas constant

T = temperature

Since we're considering the same container with the same amount of gas, the number of moles (n) and the ideal gas constant (R) remain constant. Therefore, we can rewrite the equation as:

P₁V₁/T₁ = P₂V₂/T₂

Where:

P₁ = initial pressure

V₁ = initial volume (250 mL)

T₁ = initial temperature (25 °C + 273.15 K)

P₂ = final pressure (9.96 inHg)

T₂ = final temperature (3.7 °C + 273.15 K)

Now, let's substitute the values into the equation and solve for V₂ (the final volume):

(30.07 inHg) * (250 mL) / (25 °C + 273.15 K) = (9.96 inHg) * V₂ / (3.7 °C + 273.15 K)

Simplifying the equation:

(30.07 inHg) * (250 mL) * (3.7 °C + 273.15 K) = (9.96 inHg) * V₂ * (25 °C + 273.15 K)

Now we can solve for V₂:

V₂ = (30.07 inHg * 250 mL * (3.7 °C + 273.15 K)) / (9.96 inHg * (25 °C + 273.15 K))

Calculating the expression:

V₂ ≈ 221.96 mL

Therefore, the volume of the container in the new conditions would be approximately 221.96 mL.