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OlgaM077
14 days ago
11

the image above shows a chamber with a fixed volume filled with gas at a pressure of 1560 mmHg and a temperature of 445.0 K. If

the temperature drops to 312.0K what is the new pressure of the gas in the chamber
Chemistry
1 answer:
VMariaS [1K]14 days ago
7 0

Answer:

The new gas pressure within the chamber registers at 1,093.75 mmHg

Explanation:

The Gay-Lussac Law establishes a relationship between a gas's pressure and temperature when volume remains constant. This principle asserts that gas pressure is directly tied to its temperature: as temperature increases, pressure rises, and conversely, as temperature falls, pressure also diminishes. Therefore, the Gay-Lussac law can be depicted mathematically as:

\frac{P}{T} =k

Given an initial and final state of gas, we can apply the following formula:

\frac{P1}{T1} =\frac{P2}{T2}

In this scenario:

  • P1= 1560 mmHg
  • T1= 445 K
  • P2=?
  • T2= 312 K
<psubstituting:>

\frac{1560 mmHg}{445 K} =\frac{P2}{312 K}

Calculating:

P2=\frac{1560 mmHg}{445 K} *312K

P2=1,093.75 mmHg

The new gas pressure inside the chamber is 1,093.75 mmHg

</psubstituting:>
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