Q. 33 a) State the Law of Conservation of Momentum
conserved during a collision
b) Derive a mathematical expression to show how le i
Answers
a) For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.
b) ★ Charles' Law :
It is a law stating that the volume of an ideal gas at constant pressure is directly proportional to the absolute temperature.
★ Derivation :
according to law ,
V \propto TV∝T (Pressure is constant)
We know that, when proportionality is removed then a constant is added . (K is the constant)
V = KT
{\purple{\boxed{K = \frac{V}{T}}}}
K=
T
V
constant always remains same.
for 2 gases ,
1^{st}\: gas - volume = V_{1} , temperature = T_11
st
gas−volume=V
1
,temperature=T
1
2^{nd} \: gas - volume = V_{2} , temperature = T_22
nd
gas−volume=V
2
,temperature=T
2
K = \frac{V_1}{T_1}K=
T
1
V
1
K = \frac{V_2}{T_2}K=
T
2
V
2
{\pink{\boxed{\therefore \: \frac{V_1}{T_1} = \frac{V_2}{T_2}}}}
∴
T
1
V
1
=
T
2
V
2
hence , derived .
★ Additional Information :
Formula of Gay - lussac law -
{\pink{\boxed{\frac{P_1}{T_1} = \frac{P_2}{T_2} \: (Volume\: is \: constant)}}}
T
1
P
1
=
T
2
P
2
(Volumeisconstant)
Formula of Boyle's Law -
{\pink{\boxed{\frac{P_1}{d_1} = \frac{P_2}{d_2} \: (Temperature \: is \: constant)}}}
d
1
P
1
=
d
2
P
2
(Temperatureisconstant)
Ideal gas equation :
{\pink{\boxed{PV = nRT}}}
PV=nRT
P - pressure
V = volume
n = number of moles
T = temperature
R = Ideal gas constant
\left(R = 0.0821 \: \frac{L . atm}{mol.K}\right)(R=0.0821
mol.K
L.atm
hope it helps
and please mark as brainliest.
Newton's third law states that for a force applied by an object A on object B, object B exerts back an equal force in magnitude, but opposite in direction. This idea was used by Newton to derive the law of conservation of momentum. ... B=m_{2}(v_{2}-u_{2}) (change in momentum of particle B)