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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

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Answer:

b. 1150 mmHg

General Formulas and Concepts:

Chemistry - Gas Laws

Gay Lussac Law - [tex]\frac{P_1}{T_1} =\frac{P_2}{T_2}[/tex]

  • P₁ is Pressure 1
  • T₁ is Temperature 1 in Kelvin
  • P₂ is Pressure 2
  • T₂ 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

  1. Substitute:                    [tex]\frac{1064 \ mmHg}{285.15 \ K} =\frac{P_2}{308.15 \ K}[/tex]
  2. Cross-multiply:            [tex](1064 \ mmHg)(308.15 \ K)=P_2(285.15 \ K)[/tex]
  3. Multiply:                       [tex]327872 \ mmHg \cdot K=P_2(285.15 \ K)[/tex]
  4. Isolate P₂:                    [tex]\frac{327872 \ mmHg \cdot K}{285.15 \ K}=P_2[/tex]
  5. Divide:                         [tex]1149.82 \ mmHg=P_2[/tex]
  6. Rewrite:                       [tex]P_2=1149.82 \ mmHg[/tex]

Step 5: Check

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

1149.82 mmHg ≈ 1150 mmHg

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