In this assignment, you will have 3 chances for each question. You may assume th

Question

In this assignment, you will have 3 chances for each question. You may assume that all gases in these problems are ideal gases, and remain ideal at all temperatures and pressures. For the gas constant, use R = 8.314 J/mol-K. For Avogadro's number, use Nbar 022x1023 molecules .For this assignment, you submit answers by question parts. The number of submissions remaining for each question part only changes if you submit or change the answer. Your best submission for each question part is used for your score. Be careful of units! All the conversions you need are in the information section of the lab. Suppose the volume of the syringe is 36.9 cc. You measure the pressure as 2.01 atm, and the temperature as 17.1degree C. Assume: the diameter of the plunger on the syringe is 2.32 cm. and - the number of moles of gas in the syringeoles n = - the number of molecules of gas in the syringe. N = molecules - the mass of air in the syringe. (Use the molar mass of air given in the information section of the lab.)kg Supose you push down the syringe so the volume is now 33.9 cc, with the temperature remaining unchanged. Find: - the new pressure inside the syringe.atmP = - the magnitude of the force exerted by the gas on the plunger of the syringe. F - N Suppose you suddenly press down on the syringe so the volume is now 17.6 cc, and plunge it into cold water. When the temperature is 5.6degree C, find the pressure inside the syringe. P = atm

Explanation / Answer

Let, R =8.314 J/mol-k , No = 6.022 * 10^ 23 molecules

Ideal ga equation is PV=nRT

1 atm =101325 pa

1 cc= 1/1000000 m3

1 oC = 273 o K

P = 2.01 *101325 Pa =203663.25 Pa

V= 36.9/1000000 = 0.0000369 m3

T= 17.1 +273=290.1 o K

a ) n= PV/RT =(203663.25*0.0000369 )/(8.314*290.1)=0.0031158 moles

b) N= n*No = 0.0031158 * 6.022*10^23 =1.8763 *10^21 molecules

c) mass of air = number of moles * molar mass =0.0031158 * 28.97 gm/mol =0.090264726 gm = .09026 *10^-3 kg

d) Now , V=33.9 cc=33.9/1000000 =0.0000339 m3

P=nRT/V = 0.0031158 * 8.314 * 290.1 /0.0000339 =221680.5670 Pa =221680.5670/101325=2.1878 atm

e ) F= P /A = 221680.5670/0.0116 =19110393.706 Pa/m2 N

f) Let,V= 17.6cc=17.6/1000000 = 0.0000176 m3

T= 5.6 o C =5.6+273 = 278.6 o K

P= nRT /V =0.0031158 *8.314 *278.6/0.0000176 = 410060.59490 Pa=410060.59490/101325 =4.0469 atm

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