Ideal Gas Law

Remember back 5 rows of even side are no seating zones Please turn off all downloads. You can usually pause them by clicking on them and clicking pause. Facebook and youtube are officially banned during class time (this is not me micromanaging, it’s a legit internet issue as told to me by OIT). Visit: https://checkin.ics.uci.edu/ Log in and select Chem 1A. When prompted, type the word of the day: gas Ensure when asked if you will share your location you select “allow”. Visit: https://learningcatalytics.com/ Sign in MasteringChemistry account name When prompted, type session ID:16790868 Remember back 5 rows of even side are no seating zones Visit: https://checkin.ics.uci.edu/ Word of the day: gas LearningCatalytics: session ID:16790868

Quiz Question 1 When filling into the ideal gas law PV=nRT. (not in ratio form) Which of following is true: For R use 8.314 𝐽 𝑚𝑜𝑙∗𝐾 For R use 0.0821 𝐿∗𝑎𝑡𝑚 𝐾∗𝑚𝑜𝑙 The units for temperature must be in Celsius. The units for volume must be in L The units for pressure must be in atm Visit: https://checkin.ics.uci.edu/ Word of the day: gas LearningCatalytics: session ID:16790868

Quiz Question 2 𝑃1𝑉1 𝑛1𝑇1 = 𝑃2𝑉2 𝑛2𝑇2 When using the ratio form (shown below) of the ideal gas law what is true about the units. You MUST use L for volume and atm for pressure. For n you can use number of atoms/molecules or mols For temperature you can use Celsius or Kelvin. For temperature you must always use Kelvin. 𝑃1𝑉1 𝑛1𝑇1 = 𝑃2𝑉2 𝑛2𝑇2 Visit: https://checkin.ics.uci.edu/ Word of the day: gas LearningCatalytics: session ID:16790868

Ideal Gas Law Visit: https://checkin.ics.uci.edu/ Word of the day: gas LearningCatalytics: session ID:16790868 What volume does 12.5g of argon gas at a pressure of 1.05 atm and a temperature of 322K occupy? Would the volume be different if the sample were 12.5 g of helium (under identical conditions)? 𝑃𝑉=𝑛𝑅𝑇 𝑉= 𝑛𝑅𝑇 𝑃 = 12.5𝑔 1 𝑚𝑜𝑙 39.95 𝑔 ∗0.0821 𝐿∗𝑎𝑡𝑚 𝑚𝑜𝑙∗𝐾 ∗322𝐾 1.05 𝑎𝑡𝑚 𝑉=7.87 𝐿 𝐴𝑟𝑔𝑜𝑛 𝑌𝑒𝑠 𝑖𝑡 𝑤𝑜𝑢𝑙𝑑 𝑏𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡. 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠

𝑃2=440𝑚𝑚𝐻𝑔 Ideal Gas Laws R= 𝑃1𝑉1 𝑛1𝑇1 = 𝑃2𝑉2 𝑛2𝑇2 𝑃𝑉=𝑛𝑅𝑇 A sample of gas has an initial volume of 5.6 L at a pressure of 735 mmHg. If the volume of the gas is increased to 9.4 L, what is the pressure? R= 𝑃1𝑉1 𝑛1𝑇1 = 𝑃2𝑉2 𝑛2𝑇2 𝑃𝑉=𝑛𝑅𝑇 735𝑚𝑚𝐻𝑔∗5.6𝐿 =𝑃2∗9.4𝐿 𝑃2= 735𝑚𝑚𝐻𝑔∗5.6𝐿 9.4𝐿 𝑃2=440𝑚𝑚𝐻𝑔 Visit: https://checkin.ics.uci.edu/ Word of the day: gas LearningCatalytics: session ID:16790868

Ideal Gas Law P2=1.10atm*(712+273)/298=3.63atm A 250. mL aerosol can at 25 C and 1.10 atm was thrown into an incinerator. When the temperature in the can reached 712 C, it exploded. What was the pressure in the can just before it exploded, assuming it reached the maximum pressure possible at that temperature (aka, assuming its an ideal gas)? P2=1.10atm*(712+273)/298=3.63atm Visit: https://checkin.ics.uci.edu/ Word of the day: gas LearningCatalytics: session ID:16790868

Ideal Gas Laws Yes it exceeded it. 𝑃1𝑉1 𝑛1𝑇1 = 𝑃2𝑉2 𝑛2𝑇2 Note use this for help with 11.48 homework An automobile tire has a maximum rating of 38.0 psi (gauge rating). The tire is inflated while cold to a volume of 11.8 L and a gauge pressure of 36.0 psi (gauge rating) at a temperature of 12.0 oC. When the car is driven on a hot day, the tire warms up to 65.0 oC, and its volume expands to 12.2 L. Does the pressure in the tire exceed its maximum rating? *note gauge rating is the difference between the total pressure and atmospheric pressure. Assume the atmospheric pressure on these days is 14.7 psi. 𝑃1𝑉1 𝑛1𝑇1 = 𝑃2𝑉2 𝑛2𝑇2 𝑃 2𝑡𝑜𝑡 = 𝑃1𝑉1𝑇2 𝑇1𝑉2 = 𝑃 2𝑡𝑜𝑡 = (36.0𝑝𝑠𝑖+14.7)∗11.8𝐿∗ 273+65 𝐾 273+12 𝐾(12.2𝐿) = 𝑃 2𝑡𝑜𝑡 =58.2𝑝𝑠𝑖 Yes it exceeded it. 𝑃 2𝑔𝑢𝑎𝑔𝑒 =58.2𝑝𝑠𝑖−14.7=43.5𝑝𝑠𝑖