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2 years ago

The volume of the vessel is reduced twice,and the number of molecules is reduced three times.How the pressure will change gas?

Physics

1 answer:

Yuri [45]2 years ago

4 0

Answer:

P₁ = ⅔ P₀

Explanation:

For this exercise we can use the ideal gas equation

PV = n R T

let's write the equation for the initial point, where we will use the subscript o and the end with subscript 2

Suppose the temperature does not change in the process

P₀V₀ / n₀ = P₁V₁ / n₁

They tell us that

V₁ = V₀ / 2

n₁ = n₀ / 3

we substitute

$P_o \frac{V_o}{n_o} = P_1 \frac{V_o}{2} \frac{3}{n_o}$

P₀ = $\frac{3}{2}$ P₁

P₁ = ⅔ P₀

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The volume of the vessel is reduced twice,and the number of molecules is reduced three times.How the pressure will change gas?​

The volume of the vessel is reduced twice,and the number of molecules is reduced three times.How the pressure will change gas?