a sample of argon gas is cooled and its volume went from 380. mL to 250. mL. If its final temperature was -55.0C, what was its original temperature

The initial temperature was 58.4°C

Explanation:

Given the following data:

initial volume = V₁ = 380 mL = 0.38 L

final volume = V₂ = 250 mL = 0.25 L

final temperature = T₂ = -55°C = 218 K

According to Charles's law, the volume of a gas is directly proportional to the temperature (in Kelvin). The mathematical expression is:

V₁/T₁= V₂/T₂

So, we calculate the initial temperature (V₁) as follows:

T₁ = T₂/V₂ x V₁ = 218 K/(0.25 L) x 0.38 L = 331.36 K ≅ 331.4 K

Finally, we convert the initial temperature from K to °C:

T₁= 331.4 K - 273 = 58.4°C

1. they vary with a higher number because boiling is way hotter than melting