Question

A sample of a gas occupies 1.80 L at -10. °C and 450. Torr. What will be the temperature if the pressure is increased to 800. Torr and the volume is decreased by 1/3?

Answer 1

Answer: 156 K

Step-by-step explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

`(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`

where,

`P_1` = initial pressure of gas = 450 torr

`P_2` = final pressure of gas = 800 torr

`V_1` = initial volume of gas = 1.80 L

`V_2` = final volume of gas =
`(1)/(3)* 1.80L=0.60L`

`T_1` = initial temperature of gas =
`-10^oC=273-10=263K`

`T_2` = final temperature of gas = ?

Now put all the given values in the above equation, we get:

`(450* 1.80)/(263)=(800* 0.60)/(T_2)`

`T_2=156K`

Thus the final temperature will be 156 K