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 3
5 °C?
O a. 985 mmHg
O b. 1150 mmHg
O c. 980 mmHg
d. 1.1 x 10^3 mmHg
1 answer:
Answer:
b. 1150 mmHg
General Formulas and Concepts:
<u>Chemistry - Gas Laws</u>
Gay Lussac Law - 
- P₁ is Pressure 1
- T₁ is Temperature 1 in Kelvin
- P₂ is Pressure 2
- T₂ is Temperature 2 in Kelvin
Explanation:
<u>Step 1: Define</u>
P₁ = 1.40 atm
T₁ = 12°C
P₂ = unknown
T₂ = 35°C
<u>Step 2: Identify Conversions</u>
1 atm = 760 mmHg
K = °C + 273.15
<u>Step 3: Convert</u>
P₁ = 1.40 atm = 1064 mmHg
T₁ = 12°C = 285.15 K
T₂ = 35°C = 308.15 K
<u>Step 4: Find P₂</u>
- Substitute:

- Cross-multiply:

- Multiply:

- Isolate P₂:

- Divide:

- Rewrite:

<u>Step 5: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
1149.82 mmHg ≈ 1150 mmHg
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