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

Answer:

6.30 L

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

$\frac{P_{1}\times V_{1}}{T_{1}} = \frac{P_{2}\times V_{2}}{T_{2}}$

$V_{2} = \frac{P_{1}}{P_{2}}\times\frac{T_{2}}{T_{1}}\times V_{1}$

By substituting the values, we get

V2 = 6.30 L