Question

What volume of methane gas at 237 K and 101.33 kPa do you have when the volume is decreased to 0.50 L, with a temperature of 300 K and a pressure of 151.99 kPa

Answer 1

Answer: A volume of 0.592 L of methane gas is required at 237 K and 101.33 kPa when the volume is decreased to 0.50 L, with a temperature of 300 K and a pressure of 151.99 kPa.

Step-by-step explanation:

Given:
`T_(1)` = 237 K,
`P_(1)` = 101.33 kPa,
`V_(1)` = ?

`T_(2)` = 300 K,
`P_(2)` = 151.99 kPa,
`V_(2)` = 0.50 L

Formula used is as follows.

`(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))`

Substitute the values into above formula as follows.

`(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))\\(101.33 kPa * V_(1))/(237 K) = (151.99 kPa * 0.50 L)/(300 K)\\V_(1) = 0.592 L`

Thus, we can conclude that a volume of 0.592 L of methane gas is required at 237 K and 101.33 kPa when the volume is decreased to 0.50 L, with a temperature of 300 K and a pressure of 151.99 kPa.