Question

At 300K, the pressure inside a gas-filled container is 500 kPa. If the pressure inside decreases to 100 kPa, but the volume remains the same, what is the final temperature of the gas? Answers: A.) 240 K B.) 180 K C.) 30 K D.) 60 K

Answer 1

PV = nRT

⇒P1/T1 = P2/T2
⇒500/300 = 100/T2
⇒ T2 = 100×300/500 = 60K

Answer is D

Answer 2

Answer:

The final temperature of the gas is 60K

Explanation:

The gas law to be used here is Gay-Lussac's law which states that the pressure of a given mass of gas is directly proportional to its temperature in kelvin provided volume remains constant. The law can be represented with the formula below

P₁/T₁ = P₂/T₂

where P₁ is the initial pressure (500 kPa)

T₁ is the initial temperature (300K)

P₂ is the final pressure (100 kPa)

T₂ is the final temperature (unknown)

Hence,

500/300 = 100/T₂

T₂ = $\frac{300 * 100}{500}$

T₂ = 60K

Hence, the final temperature of the gas is 60K