Question

A gas occupies 0.67 L at 350 K. What temperature is required to reduce the volume by 45%?

Answer 1

Answer: 193.3 K

Explanation:

Charles' Law: This law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

`V\propto T` (At constant pressure and number of moles)

`(V_1)/(T_1)=(V_2)/(T_2)`

where,

`V_1` = initial volume of gas = 0.67 L

`V_2` = final volume of gas =
`0.67-(45)/(100)* 0.67=0.37L`

`T_1` = initial temperature= 350 K

`T_2` = final temperature = ?

Now put all the given values in the above equation, we get the final pressure of gas.

`(0.67)/(350)=(0.37)/(T_2)`

`T_2=193.3K`

Therefore, the temperature required would be 193.3 K