Chemistry: You fill a tank with gas at 60 C to 100 kpa and seal it you decrease the temperature to 10

You fill a tank with gas at 60 C to 100 kpa and - id601005631415, angelapegues20097

Ответ:

jzane1182

21.01.2022

The given question is incomplete. The complete question is :

You fill a tank with gas at 60C and 100 kPa and seal it. You decrease the temperature to 10C but keep the volume constant. What happens to the pressure?

Thus the pressure will decrease to 85.0 kPa

Explanation:

Gay-Lussac's Law : It is defined as the pressure is directly proportional to the temperature of the gas at constant volume and number of moles.

(Volume, moles of gas are constant)

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=100kPa\\T_1=60^oC=(60+273)K=333K\\P_2=?\\T_2=10^0C=(10+273)K=283K$

Putting values in above equation, we get:

$\frac{100}{333}=\frac{P_2}{283}\\\\P_2=85.0kPa$

Thus the pressure will decrease to 85.0 kPa

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Ответ:

haydenbq

19.06.2020

Chemistry

This number can be rewritten as 650000.
Notice where the decimal is. To put this into scientific notation, move the decimal left so that it is before the 6. Count how many times the decimal moves left in order to get to the six. Since the decimal has to move left 5 digits, that is how much you have to raise 10 to. So this can be rewritten as 6.5x10^5

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You fill a tank with gas at 60 C to 100 kpa and seal it you decrease the temperature to 10

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