An ideal gas at 26.9°C and a pressure 1.60 105 Pa is in a container having a vol

Question

An ideal gas at 26.9°C and a pressure 1.60 105 Pa is in a container having a volume of 1.00 L.

(a) Determine the number of moles of gas in the container.

(b) The gas pushes against a piston, expanding to twice its original volume, while the pressure falls to atmospheric pressure. Find the final temperature. Answer in K

c) Suppose the temperature of 4.51 L of ideal gas drops from 390 K to 275 K.
If the volume remains constant and the initial pressure is atmospheric pressure, find the final pressure. Answers in Pa

d)Find the number of moles of gas.

Explanation / Answer

Part A)

From the ideal gas law, PV = nRT

We need to solve for n

n = PV/RT

V = 1 L which is .001 m3

T = 26.9oC which is 300.05 K

R is a constant = 8.31 J/mol K

So, n = (1.6 X 105)(1 X 10-3)/(8.31)(300.05)

n = 4.43 moles

Part B)

The proportionality to use here is

P1V1/T1 = P2V2/T2

(1.6 X 105)(.001)/(300.05) = (1.013 X 105)(.002)/T2

T2 = 379.9 K  

Part C)

The proportionality to use here is

P1/T1 = P2/T2

(1.013 X 105)/(390) = (P2)/(275)

P2 = 7.14 X 104 Pa

Part D)

Nack to using the ideal gas law

PV = nRT

The volume of 4.51 L is 4.51 X 10-3 m3

(1.013 X 105)(4.51 X 10-3) = n(8.31)(390)

n = .141 moles

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