25 % of energy goes into rotational energy etc., therefore 75% of the energy put in goes into Kinetic energy.

The average kinetic energy of a molecule :E

At T

total E

At T

Change in total Ek = 3/2 k T

= 6.23 x 10

This is ¾ of the total energy put in.

Total Energy put into the system = 4/3 x 6.23 x 10

Since Helium and Argon are ideal gases, the energy put into them changes the kinetic energy of the molecules.

= 3/2 k (273 + 50) x 2 x N

= 3/2 k x 323 x 2 x N

= 3/2 k (273 + 20) x 4 x N

= 3/2 k x 293 x 4 x N

Total energy = kinetic energy for Helium + kinetic energy for Argon.

= 3/2 k x 323 x 2 x N

= 3/2 k N

= 3/2 k N

For the whole system (consisting of 6 moles):total Ek = 3/2 kT N

3/2 k N

1818 = 6 T

T = 303 K

T = 30

2 moles of X at 70

All of the energy of X is Ek = 3/2 k T N

= 3/2 1.38 x 10

The total Ek in Y = 3/2 k T N = 1.38 x 10

This is only = ¾ of the total energy in Y

Therefore the total energy in Y = 4/3 x 3713 J = 4951 J

Total energy of the system = 8549 + 4951 J = 13500 J

Since, on average, the kinetic energy of each mole of the combined gas is the same (since the temperature is constant.) The kinetic energy of the system is 9/10 of the total energy of the system.

( For Y the Ek : other energies = 3 : 1. Let the Ek of a mole of Y be 3, then the Ek of each mole of X has to be 3. Then the Ek of the system = 9/10 of the total energy)

Ek = 9/10 x 13500 = 12150 = 3/2 k T N

Hence T = Ek/3/2kN = 12150/(3/2 x 1.38 x 10

2 moles of Y at 80

Kinetic Energy of Y = 3/2 k T N =3/2 k (273+ 80) 2 N

Total energy in Y = 4/3.3/2. k N

Kinetic Energy of X = 3/2k T N = 3/2 k (273+ 35) 2 N

Energy of the system = energy in Y + energy in X

= 3/2 k N

= 3/2 k N

Ek of system = 12/14 of the total energy = 12/14 x 3/2 k N

that is 3/2 k N T = 12/14 x 3/2 k N

T = 3/2 k N

The pressure at the surface equals one atmospheric pressure = 1 Atmos.

The volume at the surface V2 = 3 V1

P1 V1 = P2 V2

P1 V1 = 1 Atmos x 3 V1

P1 = 3 Atmos

but 2 Atmos of this pressure is due to the depth of water

P =r g h.

2 x 1.013 x 10

h = 20.6 m

P2 = 1 Atmos, T1 = 7 + 273 K, P1 = 3 Atmos, T2 = 27 + 273 K

P1.V1/T1 = P2.V2/T2

V2 = 3V1 x 300/280 = 3.2 V1.

P = 76 cm of Hg = 1.013 x 10

The total number of molecules remains the same:

Originally the number of molecules = N

P V = N k T or N = P V/kT

Originally N = (1.013 x 10

Finally the number of molecules are N1 and N2

Hence N = N1 + N2

Finally N1 = P V/293 k and N2 = P 2 V/373 k

therefore equating and solving for P

P = 1.15 x 10

Acetone expands by:DV = b V1 DT

= 14.87 x 10

= 89.2 x 10

= 8.9 mL

therefore the volume of air in the top = 10 - 8.9 mL = 1.1 mL

P1 = 1 Atmos

V1 = 10 mL

T1 = 20 + 273 = 293 K

P2 = ?

V2 = 1.1 mL

T2 = 32 + 273 = 305K

P1.V1/T1 = P2.V2/T2

P2 = 1 x 10 x 305/(293 x 1.1) = 9.46 Atmos

Since it is at constant pressure V1/ T1 = V2 /T2

at 90

at 120

In K

V1/ T1 = V2 /T2

30/x = 45/(x + 30)

(x + 30 since a change of 1

therefore 30x + 900 = 45x

900 = 15x or x = 60

check: 30/60 = 45/90

therefore x = 60K

therefore 90

and 120

therefore