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## Q. 8.6

Using Gay-Lussac’s Law: Finding Pressure at a Given temperature
What does the inside pressure become if an aerosol can with an initial pressure of$4.5 ×10 ^5$ Pa is heated in a fire from room temperature (20 °C) to 600 °C?

ANALYSIS This is a Gay-Lussac’s law problem because the pressure and temperature of the gas inside the can change while its amount and volume remain constant. We know three of the four variables in the equation for Gay-Lussac’s law and can find the unknown by substitution and rearrangement.

BALLPARK ESTIMATE Gay-Lussac’s law states that pressure is directly proportional to temperature. Since the Kelvin temperature increases approximately threefold (from about 300 K to about 900 K), we expect the pressure to also increase by approximately threefold, from  $4.5 × 10^{5}Pa$ to about $14 × 10^{5}Pa$ .

## Verified Solution

STEP 1: Identify known information. Of the four varables in Gay-Lussac’s law, we know $P_{1},T_{1}$ and $T_{2}$ . (Note that T must be in kelvins.)

$\begin {matrix} P_{1} = 4.5 × 10^{5}Pa \\ T_{1} = 20 °C = 293 K \\ T_{2} = 600 °C = 873 K \end {matrix}$

STEP 2: Identify answer and units.

$P_{2} = ?? Pa$

STEP 3: Identify equation. Substituting the known variables into Gay-Lussac’s law, we rearrange to isolate the unknown.

$\frac {P_{1}}{T_{1}}= \frac {P_{2}}{T_{2}} ⇒ P_{2} = \frac {P_{1}T_{2}}{T_{1}}$

STEP 4: Solve. Substitute the known information into Gay-Lussac’s law; check to make sure units cancel.

$P_{2} = \frac {P_{1}T_{2}}{T_{1}} = \frac {(4.5 ×10^{5} Pa)(378 \cancel {K})}{293 \cancel {K}} = 13 × 10 ^{5} Pa$

BALLPARK CHECK Our estimate was $14 × 10^{5} Pa$ .