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Combined Gas Law 13)a) A sample of neon gas occupies a volume of 1.00L at 27.00C and 0.660 atm. What is the new volume if the temperature is changed 60.0C and pressure is increased to 0.920atm?...

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Combined Gas Law
13)a) A sample of neon gas occupies a volume of 1.00L at 27.00C and 0.660 atm. What is the
new volume if the temperature is changed 60.0C and pressure is increased to 0.920atm?
b)750.0ml of gas at 0.395atm of pressure and at 50.00C is heated until the volume of gas is
2000.0mL and the pressure is 0.921atm. What is the final temperature?
c) A sample of neon gas occupies a volume of 2.00L at 20.00C and 0.920 atm. It is compressed
to 0.100Liters and the temperature lowered to -150.0C.What is the new pressure in atm?
Ideal Gas Law (Use R=0.0821 liter• atm/mole K)
Extracted text: Combined Gas Law 13)a) A sample of neon gas occupies a volume of 1.00L at 27.00C and 0.660 atm. What is the new volume if the temperature is changed 60.0C and pressure is increased to 0.920atm? b)750.0ml of gas at 0.395atm of pressure and at 50.00C is heated until the volume of gas is 2000.0mL and the pressure is 0.921atm. What is the final temperature? c) A sample of neon gas occupies a volume of 2.00L at 20.00C and 0.920 atm. It is compressed to 0.100Liters and the temperature lowered to -150.0C.What is the new pressure in atm? Ideal Gas Law (Use R=0.0821 liter• atm/mole K)
Answered 183 days After Jun 06, 2022

Solution

Aditi answered on Dec 06 2022
33 Votes
SOLUTION
13. a. 719 mL
.
T 2 = ( 700 m Hg ) ( 2000 m L ) ( 323 K ) ( 300 mm Hg ) ( 750 m L ) T 2 = 2010 K T 2 = 1737 ° C
c.
We have 
P1 = 0.92 atm
V1 = 2 L
T1 = 20C = (273 + 112) = 293 K
and
V2 = 0.1 L
T2 = -150C = (273 – 150)= 123K
P2 = ?
according to combined gas laws,
P1 * V1 /...
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