gaseous state easy mcq
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Originally posted by Bigcable22:
Solution :
If these units (Pressure in Pa; Volume in m3) are used, the value of the gas constant R is 8.314.
If these units (Pressure in atm; Volume in dm3) are used, the value of the gas constant R is 0.08206.
pV = nRT
pV = (m/Mr)(RT)
Mr = (m)(R)(T) / (pV)
Mr = (m)(0.08206)(T) / (pV/1000)
Mr = (m)(82.06)(T) / (pV)
Mr = (m)(22400/273)(T) / (pV)
Mr = (m)(22400)(T) / (pV)(273)
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i wasnt taught "If these units (Pressure in atm; Volume in dm3) are used, the value of the gas constant R is 0.08206."
so if i were to use to 8.314 method how do i use it?
pV=(mRT/Mr)
Pressure = p
Volume = v cm^3 = VX10^-6 m^3
m=m grams
R = 8.314 J^-1 K^-1
Temperature = T Kelvin
Standard 22.4dm^3 but how is it related to this question? ><
are my values correct of it cannot be done this way?
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Make R the subject and obtain an expression for it, knowing that when for n = 1, p = 1atm, T = 273K (ie. 0 deg C), V = 22.4dm3.
Then, proceed from there as per normal (ie. subtitute in this R expression obtained, into the equation pV = nRT. Make Mr the subject and manipulate until you reach the required answer).
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hell i used to do these mcq in 2mins..siannzz after red alert 3 and gta IV during holidays like that liao..zzz
i got another question...
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Originally posted by Bigcable22:
hell i used to do these mcq in 2mins..siannzz after red alert 3 and gta IV during holidays like that liao..zzz
i got another question...
Hint : Only one of these 4 gases is heavier (hence denser) than nitrogen gas.
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no need to apply formula such as pV=nRT then ==> Mr= conc. X [(R X T /Pressure)] to calculate?
but despite applying formula still cant get..
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Originally posted by Bigcable22:
no need to apply formula such as pV=nRT then ==> Mr= conc. X [(R X T /Pressure)] to calculate?
but despite applying formula still cant get..
Ar = 39.9g > N2 = 28g > He = 4g, Ne = 20.2g, CH4 = 16gOnly argon Ar(g), is heavier (hence denser) than nitrogen gas N2(g).
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ok thanks. i think too much XD