sapling: a sample of an ideal gas has a volume of 3.35L at 14.60 C and 1.50 ATM.

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sapling: a sample of an ideal gas has a volume of 3.35L at 14.60 C and 1.50 ATM. what is the volume of the gas at 19.20 C and 0.993 atm College of the Mainland X e Start. Here-CHEM-1406-1... X College of the Mainland x Search Q Wibiscms/mod/flcn/view.phphid 3285383 Assignment Score 79.1% Question 10 of 16 A mixture of He, Ne, and Ar has a pressure of 11.5 atm at 28.0 °C. If the partial pressure of e is 2.15 atm and the H partial pressure of Ar is 2.50 atm, what is the partial pressure of Ne? atm Check Anse

Explanation / Answer

Asample of an ideal gas has a volume of 3.35L at 14.60 C and 1.50 ATM. what is the volume of the gas at 19.20 C and 0.993 atm

From ideal gas equation, PV = nRT

                   The number of moles(n) remains constant andUniversal gas constant (R) is constant

                               SO, PV/T = constant

                  This can be written as :       P1V1/T1 = P2V2/T2

We are given that P1 = 1.50 atm ; V1 = 3.35 L ;T1 = 14.60 C= 14.6+273 = 287.6 K (conversion into Kelvin)

                             P2 = 0.993 atm ; V2 =? ; T2 = 19.20 C = 19.2+ 273 =292.2 K

                     Using P1V1/T1 = P2V2/T2 and substituting for all these terms,

                         We get V2 = P1V1T2/(T1P2)

                                         V2 = 5.14 L

The temperature in C can also be used as such.

b) Total pressure of the system of He,Ne and Ar = 11.5 atm

Partial pressure of Ar = 2.5 atm

Partial pressure of He = 2.15 atm ; Partial pressure of Ne = ?

Total pressure = sum of partial pressures

So 11.5 = 2.5 + 2.15 + pNe

pNe = 6.85 atm

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