A sample of krypton gas is stored in a container with a fixed volume at 20.0 atm of pressure. The temperature (in K) of the gas is halved. What is the new pressure of the gas, assuming that the amount of gas is constant

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07-01-2022
Chemistry

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A sample of krypton gas is stored in a container with a fixed volume at 20.0 atm of pressure. The temperature (in K) of the gas is halved. What is the new pressure of the gas, assuming that the amount of gas is constant

Sagot :

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Sebassandinidn Sebassandinidn · Answer 1 · 2022-01-17T04:54:06-05:00

This problem is describing a sample of krypton which undergoes a change in its temperature and pressure by holding its amount constant. As the initial pressure of the gas was 20.0 atm and the temperature is halved, one can use the Gay-Lussac's formula to prove that the result is 10 atm according to the following:

Gay-Lussac's law:

In chemistry, gas laws are used to relate changes in pressure, volume, moles and temperatures in a gas or gas mixture in agreement with the concept of ideal gases.

In this case, since the temperature and pressure are said to change, whereas the moles and volume of the gas remain unchanged, we can recall the Gay-Lussac's law as a directly proportional relationship between pressure and temperature:

[tex]\frac{P_2}{T_2} =\frac{P_1}{T_1}[/tex]

Thus, we can solve for P2 and subsequently plug in the initial pressure and the T2/T1 ratio (halved) to get:

[tex]P_2=\frac{P_1T_2}{T_1} \\\\P_2=20atm*0.5\\\\P_2=10atm[/tex]

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Sagot :

Gay-Lussac's law:

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