Homework 6 ± Core Chemistry Skill: Using the Gas Law Relating Temperature and Pr
ID: 571997 • Letter: H
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Homework 6 ± Core Chemistry Skill: Using the Gas Law Relating Temperature and Pressure 10 of 20> Part A Constants I Periodic Table Gay-Lussac's law states that the pressure of a gas is directly related to its absolute temperature, provided the volume and number of moles of the gas remain the same: (Figure 1) A sample of a gas is in a sealed container. The pressure of the gas is 485 torr, and the temperature is 34 °C . If the temperature changes to 94 ° C with no change in volume or amount of gas, what is the new pressure, Pz, of the gas inside the container? Express your answer with the appropriate units. View Available Hintts) In an ideal gas, particles are considered to interact only when they collide, with no loss in energy or speed occurring during the collision. Studying ideal gases allows for simplifications, such as Gay-Lussac's law. PValhue Units Submit Part B Figure 1 of 1 Using the same sample of gas (P 485 torr, T 34 °C), we wish to change the pressure to 4850 torr with no accompanying change in volume or amount of gas. What temperature T2, in Celsius, is needed to reach this pressure? Express your answer with the appropriate units View Available Hint(s) T-Value Units Submit Lower temperatureHigher temperatureExplanation / Answer
1)
Ideal gas equation is
PV = nRT
P = nRT/V
if n and V is not changed
P = kT
P1/T1 = P2/T2
P1 = 485torr= 0.6402atm ( 1torr = 0.00132atm)
T1 = 34°C = 307.15K (0°C = 273.15K)
T2 = 94°C = 367.15K
P2 = (P1/T1)×T2
= (0.6402atm/307.15K)×367.15K
= 0.765atm
= 579.5torr
2) P1/T1 = P2/T2
T2 = (P2 ×T1) /P1
P1 = 485torr = 0.6402atm
T1 = 34°C = 307.15K
P2 = 4850torr = 6.402
substituting the values
T2 = (6.402atm×307.15K)/0.6402atm
= 3071.5K
= 2798.4°C
3) P1/T1 =P2/T2
P2 = (P1/T1) ×T2
P1 = 1.15atm
T1 = 55°C = 328.15K
T2 =-37°C= 236.15K
substituting the values
P2 = (1.15atm/328.15K)×236.15K
= 0.828atm
4) P1/T1 = P2/T2
P2 =(P1/T1)×T2
P1 = 850mmHg = 1.122atm ( 1Hgmm = 0.00132atm)
T1 = - 65°C = 208.15K
T2 = 36°C = 309.15K
substituting the values
P2 = (1.122atm/208.15K)×309.15K
= 1.67atm
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