A sample of diborane gas (B2H6), a substance that bursts into flame when exposed to air, has a pressure of 345 torr at a temperature of -15ºC and a volume of 3.48 L. If conditions are changed so that the temperature is 36ºC and the pressure is 468 torr, what will be the volume of the sample?

The final volume is 3.07L

Explanation:

The general gas law will be used:

P1V1 /T1 = P2V2 /T2

V2 =P1 V1 T2 / P2 T1

Give the variables to the standard unit:

P1 = 345 torr = 345 /760 atm = 0.4539atm

T1 = -15°C = -15 + 273 = 258K

V1 = 3.48L

T2 = 36°C = 36+ 273 = 309K

P2 = 468 torr = 468 * 1/ 760 atm = 0.6158atm

V2 = ?

Equate the values into the gas equation, you have:

V2 = 0.4539 * 3.48 * 309 / 0.6158 * 258

V2 = 488.0877 /158.8764

V2 = 3.07

The final volume is 3.07L

2b2t

explanation:

2b2t