Asample 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 torrInitial temperature,T₁ = -15°C

But, K = °C + 273

          = -15°C + 273

          = 258 K

Initial volume, V₁ =3.48 L Final pressure, P₂ = 468 torrFinal 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

The correct answer is a

Idon’t know sorry i don’t know the answer i’m s sorry