plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz solve question no.25
YASH AHUJA , 6 Years ago
Grade 11
General Physics> plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz...
1 Answers
Arun
Last Activity: 6 Years ago
Dear Yash
According to the kinetic equation of pressure of a gas: P = r 1/3 But r = density of gas
r = density of gas = mass of gas / volume of gas
r = density of gas = mN / V
Putting the value of density (r) P = 1/3 (mN / V )
P V= 1/3 (mN) But PV = nRT
putting the value of PV, we get, nRT= 1/3 (mN) Since
number of mole (n) = molecules/Avogadro's number
number of mole (n) = N/NA Therefore, [N/NA] R T= 1/3 (mN)
NA R T= 1/3 (m)
3 [NA R] T= (m) But NA R = Boltzman's constant (K), thus, 3 K T= (m) Multiplying both sides by 1/2 (3/2) K T= (1/2) m (1/2) m = (3/2) K T But (1/2) m = average translational kinetic energy of gas molecules = (K.E)av
therefore, (K.E)av = (3/2) K T As (3/2) K is a constant term, therefore, (K.E)av = (constant) T
r = density of gas = mass of gas / volume of gas
r = density of gas = mN / V
Putting the value of density (r) P = 1/3 (mN / V )
P V= 1/3 (mN) But PV = nRT
putting the value of PV, we get, nRT= 1/3 (mN) Since
number of mole (n) = molecules/Avogadro's number
number of mole (n) = N/NA Therefore, [N/NA] R T= 1/3 (mN)
NA R T= 1/3 (m)
3 [NA R] T= (m) But NA R = Boltzman's constant (K), thus, 3 K T= (m) Multiplying both sides by 1/2 (3/2) K T= (1/2) m (1/2) m = (3/2) K T But (1/2) m = average translational kinetic energy of gas molecules = (K.E)av
therefore, (K.E)av = (3/2) K T As (3/2) K is a constant term, therefore, (K.E)av = (constant) T
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