A gas occupies 340 mL at 273 K and 610 torr. What is the final temperature when the pressure is increased to 1 atm, and the volume being held constant? 1 atm = 760 torr.

Question

Quizplus · Accepted Answer

We can use the combined gas law to solve this problem, which states that:

(P₁V₁)/T₁ = (P₂V₂)/T₂

where P₁, V₁, and T₁ are the initial pressure, volume, and temperature, respectively, and P₂ and T₂ are the final pressure and temperature, respectively.

We are given that:

V₁ = 340 mL
T₁ = 273 K
P₁ = 610 torr
P₂ = 1 atm = 760 torr
V₂ = V₁ (since the volume is held constant)

Substituting these values into the combined gas law, we get:

(610 torr)(340 mL)/(273 K) = (760 torr)(340 mL)/T₂

Solving for T₂, we get:

T₂ = (760 torr)(273 K)/(610 torr) = 340 K

Converting this temperature to degrees Celsius, we get:

T₂ = 340 K - 273 K = 67 °C

Therefore, the final temperature is 67 °C.

A Gas Occupies 340 ML at 273 K and 610