What is the final pressure, in mmHg, for a gas in an aerosol can at an initial pressure of 1.40 atm at 12°C which is heated to 35 °C? O a. 985 mmHg
O b. 1150 mmHg
O c. 980 mmHg
d. 1.1 x 10^3 mmHg

b. 1150 mmHg

General Formulas and Concepts:

Chemistry - Gas Laws

Gay Lussac Law -

P₁ is Pressure 1T₁ is Temperature 1 in KelvinP₂ is Pressure 2T₂ is Temperature 2 in Kelvin

Explanation:

Step 1: Define

P₁ = 1.40 atm

T₁ = 12°C

P₂ = unknown

T₂ = 35°C

Step 2: Identify Conversions

1 atm = 760 mmHg

K = °C + 273.15

Step 3: Convert

P₁ = 1.40 atm = 1064 mmHg

T₁ = 12°C = 285.15 K

T₂ = 35°C = 308.15 K

Step 4: Find P₂

Substitute:                    Cross-multiply:            Multiply:                       Isolate P₂:                    Divide:                         Rewrite:                      

Step 5: Check

We are given 3 sig figs. Follow sig fig rules and round.

1149.82 mmHg ≈ 1150 mmHg

The answer should be vaporization 

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