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VARVARA [1.3K]
3 years ago
14

If the moles and volume of a gas are held constant, dropping the temperature from 40 degrees Celsius to 20 degrees Celsius cause

s would cause an initial pressure of 3 atmospheres to drop to 1.5 atmospheres.
T/F
Chemistry
2 answers:
evablogger [386]3 years ago
8 0
It would cause a drop <span>but I am not sure  double check other answers </span>
mixer [17]3 years ago
5 0

Answer:

False.

Explanation:

Hello,

In this case, as the moles are held constant, one uses the Gay-Lussac's law in order to relate both the temperature and the pressure at the beginning and at the end of the experiment via the following relationship:

P_1T_2=P_2T_1

By solving for the pressure at the second (final) state we can substantiate whether the 1.5 atm are correct as shown below:

P_2=\frac{P_1T_2}{T_1}=\frac{3atm*293.15K}{313.15K}= 2.8 atm

Therefore, even when the pressure is correctly decreased, the given value is wrong as 2.8 atm is the correct final pressure, in this manner, the statement is false.

Best regards.

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Question 5 (1 point)
Marianna [84]

Answer:

P=12.16 atm

Explanation:

Using the formula of ideal gas law:

PV = nRT

P= nRT/V

 n= number of moles

 R= Avogadro constant = 0.0821

 T= Temperature in K => ºC + 273.15 K

P= (1.50 moles)(0.0821)( 296.15 K)/ 3.00L

P= 12.15

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Korolek [52]
I think thats the answer but feel free to check my work

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Answer:

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