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Chapter 13: Problem 45

Incandescent light bulbs are filled with an inert gas to lengthen the filament life. With the current off (at \(T=\) \(\left.20.0^{\circ} \mathrm{C}\right),\) the gas inside a light bulb has a pressure of \(115 \mathrm{kPa} .\) When the bulb is burning, the temperature rises to \(70.0^{\circ} \mathrm{C} .\) What is the pressure at the higher temperature?

Short Answer

Expert verified
Answer: The pressure of the gas inside the light bulb at a temperature of \(70.0^\circ C\) is 134.86 kPa.

Step by step solution

01

Write down the given information and convert to Kelvin

Initial temperature, \(T_1 = 20.0^\circ C = 293.15 \, K\) (convert to Kelvin by adding 273.15) Final temperature, \(T_2 = 70.0^\circ C = 343.15 \, K\) (convert to Kelvin by adding 273.15) Initial pressure, \(P_1 = 115 \, kPa\)
02

Use Gay-Lussac's Law formula to solve for final pressure

According to Gay-Lussac's Law, the ratio of the pressure to the absolute temperature is constant for a given amount of gas at constant volume. So we can write the formula as: $$ \frac{P_1}{T_1} = \frac{P_2}{T_2} $$ We are given the initial temperature, initial pressure, and the final temperature. We want to find the final pressure, \(P_2\). We can rearrange the formula to solve for \(P_2\): $$ P_2 = \frac{P_1 T_2}{T_1} $$
03

Plug in the given values and solve for the final pressure

Substitute the given values into the formula: $$ P_2 = \frac{(115\, kPa)(343.15\, K)}{293.15\, K} $$ Calculate the pressure: $$ P_2 = \frac{(115\, kPa)(343.15\, K)}{293.15\, K} = 134.86\, kPa $$
04

State the final answer

The pressure of the gas inside the light bulb at a temperature of \(70.0^\circ C\) (343.15 K) is 134.86 kPa.

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