Question
Thu January 03, 2013 By:

A BALLOON CONTAIN 5.41 dm cube OF HELIUM AT 24 degree celcius & 101.5 KPa. SUPPOSE THE GAS IN THE BALLOON IS HEATED TO 35 degree celcius , IF THE HELIUM PRESSURE IS NOW 102.8 KPa , WHAT IS THE VOLUME ?

Expert Reply
Sun January 13, 2013
 convert the temperatures to kelvins.Ti = (24 + 273) = 297 K
Tf = (35 + 273) = 308 K
Following is the data table
Vi = 5.41 dm3
Pi = 101.5 kPa
Ti = 297 K
Vf = ?
Pf = 102.8 kPa
Tf = 308 K
Apply both Boyle’s law and Charles’s law combined to get
Vf= Vi x (Pi/Pf) x (Tf/Ti)
Vf=5.41 dm3 *(101.5 kPa/ 102.8 kPa)(308 K/297 K)
Vf=5.54 dm3
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