Question

For many purposes we can treat propane (CH) as an ideal gas at temperatures above its boiling point of - 42°C. Suppose the temperature of a sample of propane gas is lowered from 25.0°C to - 22.0 °C, and at the same time the pressure is changed. If the initial pressure was 0.58 kPa and the volume decreased by 40.0%, what is the final pressure?

Answer 1

Answer: The final pressure is 0.81 kPa

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 = 0.58 kPa

`P_2` = final pressure of gas = ?

`V_1` = initial volume of gas = v

`V_2` = final volume of gas =
`v-(40)/(100)* v=0.6v`

`T_1` = initial temperature of gas =
`25^0C=(25+273)K=298K`

`T_2` = final temperature of gas =
`-22^0C=(-22+273)K=251K`

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

`(0.58* v)/(298)=(P_2* 0.6v)/(251)`

`P_2=0.81kPa`

The final pressure is 0.81 kPa