Question

A gas of certain mass occupies volume of 1.2 L at 37°C and 3 atm. At what temperature will the volume and pressure of this gas become one-third of their initial values?​

Answer 1

Answer: At a temperature of 34.44 K the volume and pressure of this gas become one-third of their initial values.

Step-by-step explanation:

Given :
`V_(1)` = 1.2 L,
`T_(1) = 37^(o)C = (37 + 273) K = 310 K`

`P_(1)` = 3 atm,
`V_(2) = (1.2)/(3) = 0.4 L` ,
`P_(2) = (3)/(3) = 1 atm`

Formula used to calculate the final temperature 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))\\(3 atm * 1.2 L)/(310 K) = (1 atm * 0.4 L)/(T_(2))\\T_(2) = (1 atm * 0.4 L * 310 K)/(3 atm * 1.2 L)\\= (124)/(3.6) K\\= 34.44 K`

Thus, we can conclude that the final temperature is 34.44 K.