Suppose the volume of the syringe is 27 cc. You measure the pressure as 1.37 atm
ID: 1702352 • Letter: S
Question
Suppose the volume of the syringe is 27 cc. You measure the pressure as 1.37 atm, and the temperature as 23.6o C. Assume: the diameter of the plunger on the syringe is 2.99 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 push down the syringe so the volume is now 22.8 cc, with the temperature remaining unchanged. Find:
- the new pressure inside the syringe. atm
- the force exerted by the gas on the syringe: N
Explanation / Answer
The volume of the syringe is V1 = 27 cc = 2.7*10^-5 m^3
The pressure P1 = 1.37 atm = 1.39*10^5 Pa
The temperature is T1 = 23.6 C = 296.6 K
a)
According to ideal gas equation
P1V1 = nRT1
n is number of moles of gas
R is universal gas constant = 8.31 J/molK
(1.39*10^5)(2.7*10^-5) = n(8.31)(296.6)
n = 0.0015 mol
The number of molecules is given by N = nNA
NA is the Avagadro number = 6.02*10^23
N = 9.03*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 = 0.0015*28.96*10^-3 = 4.34*10^-5 kg
b)
The final volume V2 = 22.8 cc = 2.28*10^5 m^3
As the process is isothermal
P1V1 = P2V2
(1.37 atm)(27 cc) = P2(22.8 cc)
P2 = 1.62 atm
The force exerted in the piston will be
F = P2*A
F = 1.64*10^5 Pa * 7.02*10^-4 m^2 (A = d^2/4)
F = 115.13 N
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