# Combining pV=nRT and pV=1/3Nmc^2

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#1
How do you combine pV=nRT and pV=1/3Nmc^2 to get the transitional kinetic energy of a mole of a monatomic gas (=3/2RT)??
0
3 years ago
#2
Is m mass and c velocity?
0
3 years ago
#3
nRT=Nmc^2 /3
mc^2=3nRT/N
mc^2 /2 =3nRT/2N
KE=3nRT/2N
If both N and n are 1 so KE=3RT/2
1
#4
nRT=Nmc^2 /3
mc^2=3nRT/N
mc^2 /2 =3nRT/2N
KE=3nRT/2N
If both N and n are 1 so KE=3RT/2
Oh my goodness!! That makes sense!! Thank you so much!
0
3 years ago
#5
I haven't done thermodynamics in physics yet but what does the big N stand for?
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#6
p = pressure
V = volume
n = number of moles
R = molar gas constant
T = temperature (in kelvin)
N = number of particles/molecules
m = mass
c^2 = r.m.s speed (root.mean.square speed)
0
3 years ago
#7
(Original post by Dandysaurus)
p = pressure
V = volume
n = number of moles
R = molar gas constant
T = temperature (in kelvin)
N = number of particles/molecules
m = mass
c^2 = r.m.s speed (root.mean.square speed)
So "monatomic" is literally a "gas" of one molecule XD. Yeah familiar with pv=nrt from chemistry and I actually did my chem project on the basic gas laws. I actually find thermodyamics really interesting and I can't wait to study it in uni physics.
0
3 years ago
#8
Oh wait does it mean a monatomic molecules as in a nobel gas or something?
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#9
So "monatomic" is literally a "gas" of one molecule XD. Yeah familiar with pv=nrt from chemistry and I actually did my chem project on the basic gas laws. I actually find thermodyamics really interesting and I can't wait to study it in uni physics.
Yep! You'll get on well with Thermodynamics, as its all about gas laws, Avogadro Constant, deriving and manipualting equations. Fair play for wanting to do Physics at Uni! I find it hard enough at A-level
0
3 years ago
#10
(Original post by Dandysaurus)
Yep! You'll get on well with Thermodynamics, as its all about gas laws, Avogadro Constant, deriving and manipualting equations. Fair play for wanting to do Physics at Uni! I find it hard enough at A-level
Cool looking forward to it. What do you wanna do at uni out of interest?
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#11
Cool looking forward to it. What do you wanna do at uni out of interest?
I'm hoping to do Diagnostic Radiography, so will need Physics too :/
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3 years ago
#12
(Original post by Dandysaurus)
I'm hoping to do Diagnostic Radiography, so will need Physics too :/
Well hopefully you wont have to worry about thermodynamics in the case.
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7 months ago
#13
nRT=Nmc^2 /3
mc^2=3nRT/N
mc^2 /2 =3nRT/2N
KE=3nRT/2N
If both N and n are 1 so KE=3RT/2
if n = moles and N = number of molecules, how can they both be 1 as i thought N = n x avagadros constant so surely if N = 1 then n would be 1/6.022x10^23?
0
7 months ago
#14
(Original post by tangus)
if n = moles and N = number of molecules, how can they both be 1 as i thought N = n x avagadros constant so surely if N = 1 then n would be 1/6.022x10^23?

pV=nRT and pV=1/3Nmc^2
1 mole so n =1
so pV= RT and pV = 1/3 Nmc^2
so RT = 1/3Nmc^2
3/2 RT = 1/2 Nmc^2 = N x 1/2mc^2 = N x KE of 1 molecule of gas

so transitional kinetic energy of a mole of a monatomic gas is 3/2 RT
1
7 months ago
#15
(Original post by golgiapparatus31)
pV=nRT and pV=1/3Nmc^2
1 mole so n =1
so pV= RT and pV = 1/3 Nmc^2
so RT = 1/3Nmc^2
3/2 RT = 1/2 Nmc^2 = N x 1/2mc^2 = N x KE of 1 molecule of gas

so transitional kinetic energy of a mole of a monatomic gas is 3/2 RT
I understand, many thanks.
1
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