Question

Initial volume is 2.3L, initial temperature 150K and initial pressure 15psi. Temperature doubles and pressure

is halved. The new volume is what?

Answer 1

`V_2=9.2L`

Step-by-step explanation:

Hello,

In this case, in terms of the combined ideal gas law:

`(p_1V_1)/(T_1)= (p_2V_2)/(T_2)`

The volume at the second state turns out:

`V_2=(p_1V_1T_2)/(T_1p_2)= (15psi*2.3L*(2*150K))/(150K*(15psi/2)) =9.2L`

Best regards.

Answer 2

V₂ = 9.2 L

Step-by-step explanation:

Initial volume = 2.3 L

initial temperature = 150 K

Initial pressure = 15 psi

Final temperature = 300 K

Final pressure = 7.5 psi

Final volume = ?

Solution:

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁T₂ /T₁ P₂

V₂= 15 psi × 2.3 L × 300 K / 150 K × 7.5 psi

V₂ =10350 psi .L. K / 1125 K .psi

V₂ = 9.2 L