1. yes
2. yes
3. There will be less gas when the saturated gas is dried. Subtract the vapor pressure of water at 25 C (23.8 mm Hg) from the 760 mmHg pressure to get the initial dry gas partial pressure. Since the pressure is maintained at 760 mm, that means the volume must be reduced by the fraction (760- initial water vapor pressure)/760
I have a packet of homework so I think I may post other questions that I'm stuck on later this weekend. So that's a heads up.
This packet deals with Gas Laws.
1. Given 100 ml of dry gas measured at 37° C and 760 mmHg, what would be its volume at 60°C?
I used Charle's Law and solved for V2.
So... (100/310)(333)=107.4ml
correct?
2. Given 100 ml of dry gas at 37°C and 760 mmHg, what would be its volume at 60°C and 800 mmHg?
I used the combined gas law and solved for V2.
So... (760)(100)(333)/(310)(800)=102ml
3. Given 325 ml of saturated gas at 760 mmHg and 25°C, what would be its volume if dry at the same pressure and temperature?
I got confused when I read 'if dry' Does that mean I have to subtract something out of 760mmHg?
And would I used the combined gas law and solved for V2?
Thanks!
4 answers
Can you explain to me how you got the vapor pressure of water, the 23.8mmHG?
How did you get that number?
And so the final number component is 0.96?
How did you get that number?
And so the final number component is 0.96?
You have to look up the vapor pressure. See
http://genchem1.chem.okstate.edu/1314SP06/Database/VPWater.html
for example.
Assuming that the gas was saturqated with H2O, the final volume is reduced to 96.9% of the initial volume, or 315 ml.
http://genchem1.chem.okstate.edu/1314SP06/Database/VPWater.html
for example.
Assuming that the gas was saturqated with H2O, the final volume is reduced to 96.9% of the initial volume, or 315 ml.
30 cubic meters of argon gas are kept under constant pressure. The gas is heated from 10.0 degrees Celsius to 293 degrees Celsius. What is the new volume of the gas?
2 answers