Transcript Ideal Gas Law - University of California, Irvine

IDEAL GAS LAW
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QUIZ QUESTION 1
• When filling into the ideal gas law PV=nRT. (not in ratio form) Which of following is true:
A) For R use 8.314
B)
𝐽
𝑚𝑜𝑙∗𝐾
For R use 0.0821
𝐿∗𝑎𝑡𝑚
𝐾∗𝑚𝑜𝑙
C) The units for temperature must be in Celsius.
D) The units for volume must be in L
E)
The units for pressure must be in atm
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QUIZ QUESTION 2
• When using the ratio form (shown below) of the ideal gas law what is true about the units.
A) You MUST use L for volume and atm for pressure.
B)
For n you can use number of atoms/molecules or mols
C) For temperature you can use Celsius or Kelvin.
D) For temperature you must always use Kelvin.
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𝑃1𝑉1 𝑃2𝑉2
=
𝑛1𝑇1 𝑛2𝑇2
IDEAL GAS LAW
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• 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
1 𝑚𝑜𝑙
𝐿 ∗ 𝑎𝑡𝑚
12.5𝑔
∗ 0.0821
∗ 322𝐾
𝑛𝑅𝑇
39.95 𝑔
𝑚𝑜𝑙 ∗ 𝐾
𝑉=
=
𝑃
1.05 𝑎𝑡𝑚
𝑉 = 7.87 𝐿 𝐴𝑟𝑔𝑜𝑛
𝑌𝑒𝑠 𝑖𝑡 𝑤𝑜𝑢𝑙𝑑 𝑏𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡. 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠
Ideal Gas Laws
• 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?
𝑃𝑉 = 𝑛𝑅𝑇
𝑃 1 𝑉1 𝑃 2 𝑉2
R=
=
𝑛1𝑇1 𝑛2𝑇2
735𝑚𝑚𝐻𝑔 ∗ 5.6𝐿 = 𝑃2 ∗ 9.4𝐿
735𝑚𝑚𝐻𝑔 ∗ 5.6𝐿
𝑃2 =
9.4𝐿
𝑃2 = 440𝑚𝑚𝐻𝑔
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Ideal Gas Law
• 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
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Ideal Gas Laws
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.
𝑃 1 𝑉1 𝑃 2 𝑉 2
=
𝑛1𝑇1 𝑛2𝑇2
𝑃2𝑡𝑜𝑡
𝑃 1 𝑉1 𝑇2
𝑃2𝑡𝑜𝑡 =
=
𝑇1 𝑉2
(36.0𝑝𝑠𝑖 + 14.7) ∗ 11.8𝐿 ∗ 273 + 65 𝐾
=
=
273 + 12 𝐾(12.2𝐿)
𝑃2𝑡𝑜𝑡 = 58.2𝑝𝑠𝑖
𝑃2𝑔𝑢𝑎𝑔𝑒 = 58.2𝑝𝑠𝑖 − 14.7 = 43.5𝑝𝑠𝑖
Yes it
exceeded it.