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aliina [53]
2 years ago
7

What would be the greatest effect of the finite size of molecules on the ideal gas law?

Physics
1 answer:
Musya8 [376]2 years ago
8 0

At high density of the molecules tends to be greater pressure that of molecules and at low density of the molecules tends to be less pressure that of molecules.

It is given that the molecules are in the ideal gas.

It is required to find the greatest effect of the finite size of molecules on the ideal gas law.

<h2>What would be the greatest effect of the finite size of molecules on the ideal gas law?</h2>

According to ideal gas law, at high density of the molecules tends to be greater pressure that of molecules and at low density of the molecules tends to be less pressure that of molecules.

From ideal gas equation,

PV = nRT

Where,

P - Pressure of the molecules

V - Volume of the molecules

n - Total amount of substance

R - Gas constant

T - Temperature of the molecules

As shown from above equation, both the quantities are directly proportional to each other and also ideal gas law follows the kinetic molecular theory  where it suggests that the volume of the molecules have negligible point.

Thus, at high density of the molecules tends to be greater pressure that of molecules and at low density of the molecules tends to be less pressure that of molecules.

Learn more about the ideal gas here:

brainly.com/question/8711877

#SPJ4

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An insect 5.00 mm tall is placed 20.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ
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Explanation:

For this exercise let's start by finding the focal length of the lens

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                1 / f = (1.70 -1)) 1 / ∞ - 1/13)

                1 / f = 0.0538

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Now we can use the constructor equation

             1 / f = 1 / o + 1 / i

             1 / i = 1 / f - 1 / o

              1 / i = -1 / 18.57 -1/20

               1 / i = -0.1038 cm

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For the height of the

image let's use magnification

                 m = h '/ h = - i / o

                  h ’= -h i / o

                  h ’= - 0.5 (-9.63) / 20

                  h ’= .0.24075 cm

b) we invert the lens

The focal length is

             1 / f = (1.70 -1) (1/13 - 1 / int)

              1 / f = 0.0538

             f = 18.57 cm

             1 / i = 1 / f -1 / o

             1 / I = 1 / 18.57 - 1/20

             1 / I = 3.85 10-3

             i = 259.74 cm

     

            h ’= - 0.5 259.74 / 20

             h ’= 6.4935 cm

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