Question

A 2.5 L container is filled with propane. The ambient temperature is 25°C and the

pressure in the container is about 3 atmospheres. If the temperature in the car
reaches 55°C (about 130°F), what will be the pressure in the container?
(please show work)

Answer 1

3.3 atm

Step-by-step explanation:

First we convert 25 °C and 55 °C to K:

• 25 °C + 273.16 = 298.16 K

• 55 °C + 273.16 = 328.16 K

We can solve this problem by using Gay Lussac's law, which states that at constant volume:

• T₁P₂=T₂P₁

Where in this case:

• T₁ = 298.16 K

• P₂ = ?

• T₂ = 328.16 K

• P₁ = 3 atm

We input the data:

• 298.16 K * P₂ = 328.16 K * 3 atm

And solve for P₂:

• P₂ = 3.3 atm

Answer 2

3.3atm is the new pressure of the container

Step-by-step explanation:

In the car, the temperature of the container increases from 25°C =298K to 55°C = 328K. To solve this question we must use the Boyle's law that state that the pressure of a gas is directly proportional to its absolute temperature under constant volume. The equation is:

P1T2 = P2T1

Where P is pressure and T absolute temperature of 1, initial state of the gas and 2, final state.

Replacing:

P1 = 3atm

T2 = 328K

P2 = ?

T1 = 298K

3atm*328K = P2*298K

P2 = 3.3atm is the new pressure of the container