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The pressure of a gas is reduced from 1200.0 mm Hg to 850.0 mm Hg as the volume of its container is increased by moving a piston from 85.00 mL to 350.00 mL. What would the final temperature be if the original temperature was 90.0 °C?

Question

The pressure of a gas is reduced from 1200.0 mm Hg to 850.0 mm Hg as the volume of its container is increased by moving a piston from 85.00 mL to 350.00 mL. What would the final temperature be if the original temperature was 90.0 °C?

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Solution

This problem can be solved using the combined gas law, which states that the ratio of the product of pressure and volume to temperature is constant for a given amount of gas. The combined gas law can be written as follows:

P1 * V1 / T1 = P2 * V2 / T2

Where: P1 = initial pressure V1 = initial volume T1 = initial temperature P2 = final pressure V2 = final volume T2 = final temperature

Given in the problem are: P1 = 1200.0 mm Hg V1 = 85.00 mL T1 = 90.0 °C = 90 + 273.15 = 363.15 K (converted from Celsius to Kelvin) P2 = 850.0 mm Hg V2 = 350.00 mL

We are asked to solve for T2.

Substituting the given values into the combined gas law gives:

1200 * 85 / 363.15 = 850 * 350 / T2

Solving for T2 gives:

T2 = 850 * 350 * 363.15 / (1200 * 85) = 315.15 K

Therefore, the final temperature would be 315.15 K. To convert this back to Celsius, subtract 273.15:

315.15 K - 273.15 = 42.0 °C

So, the final temperature would be 42.0 °C.

This problem has been solved

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