Volume of the air bubble V₁ = 1.0 cm³
= 10^-6 m³
Temperature, T₁ = 12^oC \(\Rightarrow\) = 273 K + 12°C
= 285 K.
Temperature,T₂ = 35^oC \(\Rightarrow\) = 273 K + 35°C
= 308 K
Pressure on bubble p₁ = Water pressure + Atmospheric pressure. 
= 4.93\(\times\)10⁵pa
\(\therefore \frac { { P }_{ 1 }{ V }_{ 1 } }{ { T }_{ 1 } } =\frac { { P }_{ 2 }{ V }_{ 2 } }{ { T }_{ 2 } } \)
\({ v }_{ 2 }=\frac { \left( { 4.93\times 10 }^{ 5 } \right) \times \left( { 1\times 10 }^{ -6 } \right) \times 308 }{ 258\times (1.01\times 10^{ 5 }) } \)
\(=\frac { 4.93\times 308\times { 10 }^{ -6 }\times { 10 }^{ 5 }\times { 10 }^{ -5 } }{ 285\times 1.01 } \)
\(=\frac { 4.93\times 308\times { 10 }^{ -6 } }{ 285\times 1.01 } =\frac { 1.518.44{ \times 10 }^{ -6 } }{ 287.85 } \)
= 5.275\(\times\)10^-6 m³
\(\therefore\)V₂ = 5.3\(\times\)10^-6m³

Using Graham's law of diffusion
\(\frac { { R }_{ 1 } }{ { R }_{ 2 } } =\sqrt { \frac { { M }_{ 2 } }{ { M }_{ 1 } } } \)
Squaring both side, we get.
\({ \left( \frac { { R }_{ 1 } }{ { R }_{ 2 } } \right) }^{ 2 }={ \left( \sqrt { \frac { { M }_{ 2 } }{ { M }_{ 1 } } } \right) }^{ 2 }\)
\({ \left( \frac { { R }_{ 1 } }{ { R }_{ 2 } } \right) }^{ 2 }=\frac { { M }_{ 2 } }{ { M }_{ 1 } } \)
\({ M }_{ 2 }={ M }_{ 1 }{ \left( \frac { { R }_{ 1 } }{ { R }_{ 2 } } \right) }^{ 2 }\)
\(={ \left( \frac { { 28.7 } }{ { 7.2 } } \right) }^{ 2 }\)
\(=\frac { 823.69 }{ 51.84 } =15.88\)
=16
The gas is identified as oxygen.

(i) The molecules in a gas are small and very far apart. Most of the volume which a gas occupies is empty space.
(ii) Gas molecules are in constant random motion. Just as many molecule are moving in one direction as in any other.
(iii) Molecules can collide with each other and with the walls of the container. Collisions with the walls account for the pressure of the gas.
(iv) When collisions occur, the molecules lose no kinetic energy; that is, the collisions are said to be perfectly elastic. The total kinetic energy of all the molecules remains constant unless there is some outside interference with the molecules.
(v) The molecule exert no attractive or repulsive forces on one another except during the process of collision. Between collisions, they move in straight lines.

(i). It states that the rate of cooling of a body is directly proportional to the temperature differ between the body and the surroundings.
(ii) Consider a spherical calorimeter of mass m whose outer surface is blackened. It is filled with hot water of mass m, the calorimeter with thermometer is suspended from a stand.
(iii) The calorimeter & the hot water radiate heat energy to the surrounding. Using a stop clock, the temperature is noted for every 30 sec. interval of time tree the temperature falls by about 20°C. The readings are tabulated.
(iv) If the temperature falls from T₁ to T₂ in I see, the quantity of heat energy lost by radiation Q = (ms + m₁s₁)(T₁ - T₂), where 's' is the specific heat capacity of the material of the calorimeter & S₁ - specific heat capacity of water.
Rate of cooling =\(\frac{Heat energy lost}{timet_n}\)
\(\therefore \frac { Q }{ E } =\frac { \left( ms+{ m }_{ 1 }{ m }_{ 1 } \right) \left( { T }_{ 1 }+{ T }_{ 2 } \right) }{ t } \)
Room temperature - T_o
(v) Average excess temperature of the  colorimeter over that of the surroundings 
\(-\frac { { T }_{ 1 }-{ T }_{ 2 } }{ 2 } ={ T }_{ 0 }\)
(vi) Acceleration to Newton's law of cooling 
\(\frac { Q }{ T } =\left( \frac { { T }_{ 1 }+{ T }_{ 2 } }{ 2 } -{ T }_{ 0 } \right) \)
(vii) Assume the pressure of the gas remains constant during an infinitesimally small outward displacement dy then work done dW - F. dx = P.A. dx
dW = pdv
Total work done by the gas from volume
V₁ to v₂ is \(W=\int _{ { v }_{ 1 } }^{ { v }_{ 2 } }{ pdv } \)
(ix) But pv\(\gamma\) = constant (k)
\(W=\int _{ { v }_{ 1 } }^{ { v }_{ 2 } }{ k{ v }^{ \gamma } } dv=k{ \left[ \frac { { v }^{ \gamma -1 } }{ 1-\gamma } \right] }_{ { v }_{ 1 } }^{ { v }_{ 2 } }\quad \left[ \because p=\frac { k }{ { v }^{ \gamma } } \right] \)
\(\therefore W=\frac { k }{ 1-\gamma } \left[ { v }_{ 2 }^{ 1-\gamma }{ -v }_{ 1 }^{ 1-\gamma } \right] \)
\(W=\frac { 1 }{ 1-\gamma } \left[ { kv }_{ 2 }^{ 1-\gamma }{ -kv }_{ 1 }^{ 1-\gamma } \right] \)
\({ p }_{ 2 }{ v }_{ 2 }^{ \gamma }={ p }_{ 1 }{ v }_{ 2 }^{ \gamma }\)
(x) Subtract the value of k
\(\therefore W=\frac { 1 }{ 1-\gamma } \left[ { p }_{ 2 }{ v }_{ 2 }^{ \gamma },{ v }_{ 2 }^{ 1-\gamma }-{ p }_{ 1 }{ v }_{ 1 }^{ \gamma }{ v }_{ 1 }^{ 1-\gamma } \right] \)
\(W=\frac { 1 }{ 1-\gamma } \left[ { { p }_{ 2 }v }_{ 2 }^{ }{ -{ p }_{ 1 }v }_{ 1 }^{ } \right] \)
It T~ is the final temperature of the gas in adorable expansion, then
p₁v₁ = RT₁P₂v₂ = RT₂
\(\therefore W=\frac { 1 }{ 1-\gamma } \left[ { R }_{ 2 }^{ }{ -R }_{ 1 }^{ } \right] \)
This is the equation for the work done during adiabatic process.

Consider are mole of an ideal gas enclosed in a cylinder provided with a frictionless piston of area A.
P - pressure of gas
V - volume of gas
T - absolute temperature gas
dQ - quantity of heat supplied
To keep the volume of the gas constant a small W_t is placed over the piston.
The pressure and temperature increase to p + dp and T+ dt.
dQ is used to increase the internal energy dU of the gas. But the gas does not do any work [dw = 0]
\(\therefore\) dQ = dU = 1\(\times\)C_v\(\times\)dT.
Now the W_t is removed. The piston now moves upwards thus a dist. dx, the pres. of the enclosed gas equal to atmosphere pressure P. Due to expansion, temperature decreases.
Now a quantity of heat dQ is supplied till its temperature become T + \(\Delta\)T. This heat energy is not only used to increase the internal energy dU of the gas but also to do exists wor k dW in moving the piston upwards.
\(\therefore\) dQ¹ = dU+dW
At constant pressure
dQ¹ = c_p dT 
\(\therefore\) c_p dT = C_v dT + dW
work done dW = Force\(\times\)dist.
=p\(\times\)A\(\times\)dx
dW = p. dv [A. dx = dv change in volume]
\(\therefore\) c_p dT = c_v dT + pdv ........... (1)
The eqn of state of an ideal gas is
pv = RT
Difference both the sides
pdv = RdT  ............. (2)
Subtract (2) in (1)
c_pdT = c_vdT+RdT
 c_p= c_v+R
\(\therefore\) C_p-C_v + R.
This equation is known as Meyer's electron.