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Suppose the volume of the syringe is 42.5 cc. You measure the pressure as 1.9 at

ID: 2111491 • Letter: S

Question

Suppose the volume of the syringe is 42.5 cc. You measure the pressure as 1.9 atm, and the temperature as 22.1 C. Assume: the diameter of the plunger on the syringe is 2.29 cm.

a)Find
- n, the number of moles of gas in the syringe:
moles
- N, the number of molecules of gas in the syringe: molecules
- Use the molear mass of air given to calculate the mass of air in the syringe: kg

b)Supose you push down the syringe so the volume is now 34.1 cc, with the temperature remaining unchanged. Find:
- the new pressure inside the syringe. atm
- the force exerted by the gas on the plunger of the syringe: N


a) Assume that air pressure is 1.00 atm and the air temperature is 20.6 C. If the abolute zero apparatus is open to the air, how many moles of air will it hold?
moles

b) Suppose now that the apparatus is sealed, and the temperature rises to 30.1 C. Find the pressure inside the apparatus now.
atm

Explanation / Answer

The volume of the syringe is V1 = 42.5cc = 4.25*10^-5 m^3


The pressure P1 = 1.9 atm = 1.910^5 Pa

The temperature is T1 = 22.1 C = 295.1K

a)

According to ideal gas equation

P1V1 = nRT1

n is number of moles of gas

R is universal gas constant = 8.31 J/molK


(1.9*10^5)(4.25*10^-5) = n(8.31)(295.1)


n = 0.00329 mol

The number of molecules is given by N = nNA

NA is the Avagadro number = 6.02*10^23

N = 19.8*10^20 molecules

The number of mole is given by

n = m/M

m is mass of the gas

M is molecular mass of the gas = 28.96*10^-3 kg/mol

The mass of the air in the syringe is m = 0.00329*28.96*10^-3 = 9.527*10^-5 kg

b)

The final volume V2 =34.1 cc = 3.4110^5 m^3

As the process is isothermal

P1V1 = P2V2


(1.9 atm)(42.5 cc) = P2(34.1 cc)

P2 = 2.36 atm

The force exerted in the piston will be

F = P2*A

F = 2.36*10^5 Pa * 7.02*10^-4 m^2 (A = πd^2/4)

F = 166.23 N

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