Answer to Question #283134 in Physics for Kimgcash

Question

A sample of argon gas was cooled, and its volume went from 380.mℓ to 250.mℓ. If its final temperature was −45.0◦C, what was its original temperature?

Answer 1

Assuming the pressure did not change during the cooling process, we have:

$$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2},\\\space\\ T_2=\dfrac{V_2}{V_1}T_1=\frac{250}{380}(273+(-45))=150\text{ K},\\ t_2=T_2-273=-123°\text C.$$