August 04, 2020, 07:02:28 AM
Forum Rules: Read This Before Posting

### Topic: Vapour pressure  (Read 3578 times)

0 Members and 1 Guest are viewing this topic.

#### T

• Full Member
• Posts: 119
• Mole Snacks: +0/-0
##### Vapour pressure
« on: July 18, 2015, 06:37:56 PM »
Hello,

The question is:

Butane has a vapour pressure of 47.156 kPa at 253 K, and 149.03 kPa at 283 K. Is butane gaseous or non-gaseous under the following conditions:

Temp. = 253 K      Pressure = 1 atm
Temp. = 253 K      Pressure = 2 atm
Temp. = 283 K      Pressure = 1 atm
Temp. = 283 K      Pressure = 2 atm

So firstly, I have searched what vapour pressure means. And from what I read, vapour pressure is the pressure of the vapour of a liquid at equilibrium in a closed system. And can only affected by temperature.

So since butane has a vapour pressure of 47.156 kPa at 253 K, at 253 K and 1 atm the vapour pressure of butane would still be 47.156 kPa. However, I am not sure how the external pressure would affect the physical state of butane. Could someone explain this? Also does the 1 atm mean that in a closed system, the pressure in it is 1 atm?

Thanks

#### mjc123

• Chemist
• Sr. Member
• Posts: 1760
• Mole Snacks: +248/-11
##### Re: Vapour pressure
« Reply #1 on: July 18, 2015, 06:44:25 PM »
What happens when the vapour pressure of a liquid is greater than the external pressure?

#### T

• Full Member
• Posts: 119
• Mole Snacks: +0/-0
##### Re: Vapour pressure
« Reply #2 on: July 18, 2015, 07:00:45 PM »
The liquid boils doesn't it?

So when butane has:
Temp. = 283 K      Pressure = 1 atm
It would be gaseous?

And for the others it would be non-gaseous?

#### mjc123

• Chemist
• Sr. Member
• Posts: 1760
• Mole Snacks: +248/-11
##### Re: Vapour pressure
« Reply #3 on: July 19, 2015, 05:34:49 PM »
Correct

#### T

• Full Member
• Posts: 119
• Mole Snacks: +0/-0
##### Re: Vapour pressure
« Reply #4 on: July 20, 2015, 05:09:42 AM »
Thanks mjc123