Question

A sample of argon gas is cooled and its volume went from 380 mL to 250 mL. If its final temperature was -55° C, what was its original temperature?

Answer 1

1) Convert -55 °C to Kelvin and you get 273 + (-55) = 218 K . At this temperature the volume of the gas is 250 mL.

2) At Temperature `T_1` Kelvin ,this temperature the volume of the gas is 380 mL.

Remember that you have to plug into the equation in a very specific way. The temperatures and volumes come in connected pairs and you must put them in the proper place.

Using Charles law equation;

`V_1` / `T_1` = `V_2` / `T_2`

`V_1` = 380 mL , `T_1` = ?

`V_2` = 250 mL , `T_2` = 218 K

plug in the values;

380 mL /`T_1` = 250 mL / 218 K

Cross-multiply and divide:

380 mL x 218 K = 250 mL x `T_1`

82840 mL K = 250 mL x `T_1`

`T_1` = 82840 mL K / 250 mL

`T_1` = 331.36 K