Question

The pressure of 4.20 L of an ideal gas in a flexible container is decreased to one-third of its original pressure, and its absolute temperature is decreased by one-half. What is the final volume of the gas?

Answer 1

6.30 L

Step-by-step explanation:

P1 = P, V1 = 4.20 L, T1 = T

P2 = P/3, V2 = ?, T2 = T/2

Where, V2 be the final volume.

Use ideal gas equation

`(P_(1)* V_(1))/(T_(1)) = (P_(2)* V_(2))/(T_(2))`

`V_(2) = (P_(1))/(P_(2))*(T_(2))/(T_(1))* V_(1)`

By substituting the values, we get

V2 = 6.30 L