# 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?

Jun 27, 2014

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