Q. 214.0( 1 Vote )

# Figure shows a cylindrical tube of volume V with adiabatic walls containing an ideal gas. The internal energy of this ideal gas is given by 1.5 nRT. The tube is divided into two equal parts by a fixed diathermic wall. Initially, the pressure and the temperature are p_{1}, T_{1} on the left and p_{2}, T_{2} on the right. The system is left for sufficient time so that the temperature becomes equal on the two sides.

(a) How much work has been done by the gas on the left part?

(b) Find the final pressures on the two sides.

(c) Find the final equilibrium temperature.

(d) How much heat has flown from the gas on the right to the gas on the left?

Answer :

a) According to the question, the diathermic separator between both the part is fixed. So, no change in volume will be observed. And as we know work done on gas is PV. Therefore, no work will be done on the left part during the process as the volume is not changing.

First, we will calculate the final temperature and then final pressure.

(c) Given

Pressure of left chamber =P_{1}

Pressure of right chamber =P_{2}

Temperature of left chamber =T_{1}

Temperature of right chamber =T_{2}

Let the number of moles in the left chamber be n_{1}

Number of moles in the right chamber be n_{2}

Diathermic wall has divided the tube in two equal part. So, the volume of the left and the right chamber will be V/2.

Applying ideal gas equation in the left chamber,

Similarly applying ideal gas equation in the right chamber,

Total number of moles n=n_{1} +n_{2}

The internal energy of ideal gas is given as

U=nC_{v}T

Where C_{v}=molar specific heat at constant volume

T=temperature.

According to question,

U=1.5nRT

nC_{v}T=1.5nRT

So, C_{v}=1.5R

The internal energy of the left chamber U_{1}=n_{1}C_{v}T_{1}

The internal energy of right chamber U_{2}= n_{2}C_{v}T_{2}

Total internal energy U= U_{1}+U_{2}

1.5nRT= n_{1}C_{v}T_{1}+ n_{2}C_{v}T_{2}

1.5nRT=C_{v}(n_{1}T_{1}+n_{2}T_{2})

Substituting the value of C_{v}

1.5nRT=1.5R(n_{1}T_{1}+n_{2}T_{2})

nT=n_{1}T_{1}+n_{2}T_{2}

substituting the value of n_{1} and n_{2} in above equation

Substituting the value of n in the above equation,

Thus, final equilibrium temperature T=.

(b) Now we will find final pressures on both sides.

Let final pressure of left chamber P_{1}’

Final pressure of right chamber P_{2}’

Applying ideal gas equation in the left chamber before and after equilibrium

From equation (i) and (ii),

Substituting the value of T,

Applying ideal gas equation in the right chamber before and after equilibrium

From equation (iii) and (iv),

Substituting the value of T,

(d) The internal energy of ideal gas is given as

U=nC_{v}T

Where C_{v}=molar specific heat at constant volume

T=temperature.

As stated in part (a) no work will be done on either chamber of the vessel as the diathermic separator is fixed.

So, ΔW=0 for the right chamber of the tube.

From first law of thermodynamics, we know that,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

ΔQ=ΔU

Change in internal energy of the right chamber after equilibrium has reached will be

ΔU=n_{2}C_{v}T_{2} -n_{2}C_{v}T

Substituting the value of n_{2}, C_{v} and T in the above equation

Thus, heat flown from left to right chamber is

.

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