Question

An adiabatic vessel of total volume V is divided into two equal parts by a conducting separator. The separator is fixed in this position. The part on the left contains one mole of an ideal gas (U = 1.5 nRT) and the part on the right contains two moles of the same gas. Initially, the pressure on each side is p. The system is left for sufficient time so that a steady state is reached. Find (a) the work done by the gas in the left part during the process, (b) the temperature on the two sides in the beginning, (c) the final common temperature reached by the gases, (d) the heat given to the gas in the right part and (e) the increase in the internal energy of the gas in the left part.

Answer 1

(a) Since the conducting wall is fixed, the work done by the gas on the left part during the process is zero because the change in volume will be zero due to the fixed position of the wall.

(b) For left side:-

Let the initial pressure on both sides of the wall be p.

We know,

Volume = \[\frac{V}{2}\]

Number of moles, n  = 1

Let initial temperature be T₁.

Using the ideal gas equation, we get

\[\frac{PV}{2} = nR T_1 \]

\[ \Rightarrow \frac{PV}{2} = (1) RT\]

\[ \Rightarrow T_1 = \frac{PV}{(2 \text{moles}) R}\]

For right side:-

Number of moles, n = 2

Let the initial temperature be T₂.

We know,

Volume = \[\frac{V}{2}\]

\[\frac{PV}{2} = nR T_2 \]

\[ \Rightarrow T_2 = \frac{PV}{(4 \text{moles}) R}\]

(c)

Here,

U = 1.5nRT

T = temperature at the equilibrium

P₁ = P₂ = P

n₁ = 1 mol

n₂ = 2 mol

Let T₁ and T₂ be the initial temperatures of the left and right chamber respectively.

Applying eqn. of state

For left side chamber

`PV/2=n_1RT_1`

`rArrPV/2=RT_1`

`rArr PV = 2RT_1`

`rArrT_1=(PV)/(2R)`

Right side chamber

`PV/2=n_2RT_2`

`rArrPV/2=2RT_2`

`rArrT_2=(PV)/(4R)`

We know that total n = n₁ + n₂ = 3

U₁ = n₁C_vT₁ = C_vT₁

U₂ = n₂C_vT₂ = 2C_vT₂

U =nC_vT

U = U₁ + U₂

`rArr3C_"v"T=C_"v"T_1+2C_"v"T_2`

`rArr3T=T_1+2T_2`

`rArr3T=(PV)/(2R)+(2PV)/(4R)`

`rArrT=(PV)/(3R)`

(d) For RHS:-

∆Q = ∆U as ∆W = 0

∆U = 1.5 n₂R (T − T₂)

When T is the final temperature and T₂ is the initial temperature of side 1, we get

= 1.5 × 2 × R (T − T₂)

= 1.5 × 2 × \[\frac{4PV - 3PV}{4 \times 3 \text{mole}}\]

\[= \frac{3 \times PV}{4 \times 3 \text{mole}} = \frac{PV}{4}\]

(e) If dW = 0, then using the first law, we get

dQ = − dU

⇒ dU = − dQ = \[\frac{- PV}{4}\]