Question

For many purposes we can treat methane (CH4 as an ideal gas at temperatures above its boiling point of -161. °C Suppose the temperature of a sample of methane gas is raised from -19.0 °C to 12.0 °C, and at the same time the pressure is changed. If the initial pressure was 5.6 atm and the volume decreased by 50.0%, what is the final pressure? Round your answer to the correct number of significant digits.

Answer 1

Answer: The final pressure is 12.6 atm

Step-by-step explanation:

The combined gas equation is,

`(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`

where,

`P_1` = initial pressure of gas = 5.6 atm

`P_2` = final pressure of gas = ?

`V_1` = initial volume of gas = v

`V_2` = final volume of gas =
`v-(50)/(100)v=0.5v`

`T_1` = initial temperature of gas =
`-19.0^0C=(-19+273)K=254K`

`T_2` = final temperature of gas =
`12.0^0C=(12.0+273)K=285K`

Now put all the given values in the above equation, we get:

`(5.6* v)/(254)=(P_2* 0.5v)/(285)`

`P_2=12.6atm`

The final pressure is 12.6 atm