Oxygen gas is at a temperature of 40°C when it occupies a volume of 2.3 liters. To what temperature should it be raised to occupy a volume of 6.5 liters?

Jun 27, 2014

884.5 K

Explanation:

1) Convert 40 °C to Kelvin and you get 273 + 40 = 313 K .At this temperature the volume of the gas is 2.3 L.

2) At Temperature ${T}_{2}$ Kelvin ,this temperature the volume of the gas is 6.5 L.

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;

${V}_{1}$ / ${T}_{1}$ = ${V}_{2}$ / ${T}_{2}$

${V}_{1}$ = 2.3 L , ${T}_{1}$ = 313 K

${V}_{2}$ = 6.5 L , ${T}_{2}$ = ?

plug in the values;

2.3 L / 313 K = 6.5 L / ${T}_{2}$

Cross-multiply and divide:

2.3 L x (${T}_{2}$) = 6.5 L x 313 K

2.3 L x (${T}_{2}$) = 2034.5 L K

${T}_{2}$ = 2034.5 LK / 2.3 L

${T}_{2}$ = 884.5 K