Question

- In a cylindrical container with a base area A, there are N monatomic gas particles at temperature T which are ideal. The upper part of the container is closed with a lid which has a weight M and can moves upward and downward without friction . There is vacuum on the lid and the whole system is under gravity.
a) Calculate the balance position of the lid. When performing this calculation, you can assume that the cover is quite narrow in the vertical.

- Suppose the system is completely isolated from the outside, doubling the weight of the lid (2M).
b) Determine the new temperature T* of the gas and the new equilibrium position of the lid.
c) Calculate the change in the entropy of the gas.

Answer 1

Given,

Base area is A

There are N mono-atomic gas particles

Temperature is T

Now,

(a)

Let k be the Avogadro's number.

Number of moles of gas, n = N/K

Now,

Weight of the lid is M

So, pressure due to lid is, P = weight of lid / Base area

                                             = M/A

let the balance point be at the distance L from bottom

So, volume, V = L*A

Now,

Applying ideal gas law

=> P*V = n*R*T

=> (M/A)*(A*L) = (N/K)*R*T

=> M*L = (N*R*T)/(K) ...........................(1)

or L = (N*R*T)/(M*K)

(b)

Now, weight becomes 2M

So, Pressure becomes, P'= (2M/A)

Since, system is isolated, Volume remains the same,

Let the new temperature be T'

Now,

Applying ideal gas law

=> P'*V = n*R*T'

=> (2M/A)*(A*L) = (N/K)*R*T

=> 2M*L = (N*R*T')/(K)

=> T' = (2*M*L*k)/(N*R)

From equation(1), we can see that

=> T = (M*L*K)/(N*R)

Hence,

=> T' = 2*T

Position of lid remains the same.

(c)

Change in internal energy, U = (3/2)*k*(T'-T)

=> U = (3/2)*k*(2T-T) = U = (3/2)*k*T

Now,

Average temperature, T'' = (T + T')/2 = (2T + T)/2

                                         = (3*T)/2

Now, we know

Change in entropy, S = Q/T

=> S = {(3/2)*k*T}/{(3*T)/2}

         = k = n*R

         = (N*R)/K