Question

What temperature (in °C) did an ideal gas shift to if it was initially at -23.0 °C at 4.62 atm and 35.0 L and the pressure was changed to 8.71 atm and the volume changed to 15.0 L?

Answer 1

Answer:

-71°C

Explanation:

Let's combine the Ideal Gases Law in the two situations, as the number of moles keeps on constant. Of course R, will be the same in both cases because it is a universal physic constant

P₁ . V₁ / T₁ = P₂ . V₂ / T₂

Remember that T° must be in K (T°C + 273)

-23°C + 273 = 250K

(4.62 atm . 35L) / 250K = (8.71 atm . 15L) / T₂

0.6468 L.atm/K =  (8.71 atm . 15L) / T₂

T₂ = 8.71 atm . 15L / 0.6468 L.atm/K → 201.9K

T₂ = 202K - 273 = -71°C