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shtirl [24]
3 years ago
8

I will give Brainliest

Mathematics
2 answers:
ioda3 years ago
4 0

Answer:

The volume(V) of gas will increase when the pressure of the gas decreases assuming all the other variables are held constant.

Step-by-step explanation:

 P = Pressure of the ideal gas

            V = Volume of the ideal gas

            n = Moles of the ideal gas

            R = Universal gas constant

            T = Temperature of the ideal gas

kogti [31]3 years ago
3 0

The volume(V) of gas will increase when the pressure of the gas decreases assuming all the other variables are held constant.

Explanation:

The ideal gas equation of an ideal gas is given:

    PV=nRt        

Where:

           P = Pressure of the ideal gas

            V = Volume of the ideal gas

            n = Moles of the ideal gas

            R = Universal gas constant

            T = Temperature of the ideal gas

On rearranging the above equation:

From this expression, we can deduce to things:

The pressure of the ideal gas is directly proportional to the moles of the ideal gas and the temperature of the ideal gas.

               

The pressure of the ideal gas is inversely proportional to the volume of the ideal gas.

               

So, from this, we can conclude that volume(V) of gas will increase when the pressure of the gas decreases assuming all the other variables are held constant.

Learn more about the ideal gas equation here:

brainly.com/question/4147359?referrer=searchResults

brainly.com/question/1056445?referrer=searchResults

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let X(t) denote grams of C formed in  t mins.

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\frac{dX}{dt} \infty(40-\frac{2}{3}X)(50-\frac{1}{3}X)->\frac{dX}{dt}=k(90-X)\\\therefore \frac{dX}{(90-X)^2}=kdt->\int{\frac{dX}{(90-X)^2}} \, =\int {k} \, dt  \\\therefore \frac{1}{90-X}=kt+c->90-X=\frac{1}{kt+c}\\\\X(t)=90-\frac{1}{kt+c}

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