At -11°c a sample of carbon monoxide gas exerts a pressure of 0.37 atm. what is the pressure when the volume of the gas is reduced to one third or the original value at the same temperature?

The pressure will be 1.11 atm.

Based on the given information,  

•A sample of carbon dioxide gas exerts a pressure of 0.37 atm.

•The volume of the gas is reduced to one-third, therefore, let us consider that the initial volume of the gas is V and the final volume of the gas will be V/3.

Now based on ideal gas law,  

PV=nRT

•At constant temperature, PV will be constant.  

So, P₁V₁ = P₂V₂

Now putting the values we get,  

0.37atm×V = P₂×V/3

3×0.37 = P₂

P₂ = 1.11 atm

Thus, when the volume of the gas is reduced to one-third, the pressure of the gas will become 1.11 atm.

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The pressure will be 1.11atm

Explanation:

V1 = V

P1 = 0.37 atm

V2 = (1/3)V

P2 =?

P1V1 = P2V2

0.37 x V = P2 x (1/3)V

P2 = (0.37 x V) / (1/3)V

P2 = 0.37 x 3 = 1.11atm

"record the time to complete a chemical reaction" would most likely be part of a lab procedure.