Question

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

Answer 1

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