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

3.072 L

Explanation:

We are given;

Initial pressure, P₁ = 345 torr Initial temperature, T₁ = - 15°C

But, K = °C + 273

= - 15°C + 273

= 258 K

Initial volume, V₁ = 3.48 L Final pressure, P₂ = 468 torr Final temperature, T₂ = 36°C

= 36 °C + 273

= 309 K

We are required to calculate the new volume, V₂

Using the combined gas law;

P₁V₁/T₁ = P₂V₂/T₂

We can calculate the new volume, V₂; Rearranging the formula;

V₂ = (P₁V₁T₂) : (T₁P₂)

= (345 torr * 3.48 L * 309 K) : (258 K * 468 torr)

= 3.072 L

Therefore, the new volume of the sample is 3.072 L