Combined Gas Law
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- Thecombined gas lawexpresses the relationship between the pressure, volume, and absolute temperature of a fixed amount ofgas. For acombined gas lawproblem, only the amount ofgasis held constant.
- But before that, I will discuss first the Combined gas law, and I will give examples.
- Example:A gas has a volume of 900.0 mL at −25.0 °C and 400.0 torr. What would the volume of the gas be at 220.0 °C and 500.0 torr of pressure?Given:V1= 900.0 mLT1= -25°C + 273 = 248 Kp1= 400.0 torrT2= 22.0°C + 273 = 295 KP2= 500.0 torrRequired: V2 = ?Equation:P1V1T2 = P2V2T1= V2= P1V1T2 P2T1Solution: V2 = P1V1T2 P2T1V2 =(400.0)(900.0 mL)(295) (500.0)(248)V2 = 106200000 mL 124000Answer:V2 = 856.45 mL
- Group 1BlakeAlexTaylorKennedyGroup 2AveryBrooklynCameronEzra
- I will now group you into two groups!
- ACTIVITY690.0 mL of oxygen are collected over water at 26.0 °C and a total pressure of 725.0 mm of mercury. What is the volume of dry oxygen at 52.0 °C and 800.0 mm pressure?
- Who will answer this?
- Yes, Group 1?
- 690.0 mL of oxygen are collected over water at 26.0 °C and a total pressure of 725.0 mm of mercury. What is the volume of dry oxygen at 52.0 °C and 800.0 mm pressure?Given:P1=725.0 mmHgP2=800.0 mmHgV1=690.0 mLT1 =26.0 °C + 273= 299KT2 =52.0 °C + 273= 325KRequired: V2 = ?Equation:P1V1T2 = P2V2T1= V2 = P1V1T2 P2T1Solution:V2 =P1V1T2 P2T1V2 = (725.0)(690.0mL)(325) (800.0)(299)V2 = 162581250mL 239200Answer:V2 = 679.69mL
- That's for our discussion for today. I hope that you've learned a lot about our topic.
- Prepare for our short review about Behavior of Gases and Combined gas law.
- And that's correct! Thank for your participation!
- Goodbye Ms. Sanchez!
- Goodbye Class! See you tomorrow!
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