Question

A gas is in a container with an initial volume of 2.50L, initial pressure of 1.34 atm, and initial temperature of 308 K. What would be the pressure if the gas was heated to 373 K and the volume decreased to 1.90 L? *Hint use the Combined Gas Law.

Answer 1

Answer: The pressure is 2.14 atm if the gas was heated to 373 K and the volume decreased to 1.90 L.

Step-by-step explanation:

Given:
`V_(1) = 2.50 L`,
`P_(1) = 1.34 atm`,
`T_(1) = 308 K`

`P_(2) = ?` ,
`V_(2) = 1.90 L`,
`T_(1) = 373 K`

Formula used to calculate the final pressure is as follows.

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

Substitute the values into above formula as follows.

`(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))\\(1.34 atm * 2.50 L)/(308 K) = (P_(2) * 1.90 L)/(373 K)\\P_(2) = (1.34 atm * 2.50 L * 373)/(308 K * 1.90 L)\\= 2.14 atm`

Thus, we can conclude that the pressure is 2.14 atm if the gas was heated to 373 K and the volume decreased to 1.90 L.