An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume $\mathrm{V}_{1}$ and contains ideal gas at pressure $\mathrm{P}_{1}$ and temperature $\mathrm{T}_{1}$. The other chamber has volume $V_{2}$ and contains ideal gas at pressure $P_{2}$ and temperature $T_{2}$. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be

(A) $\frac{T_{1} T_{2}\left(P_{1} V_{1}+P_{2} V_{2}\right)}{P_{1} V_{1} T_{2}+P_{2} V_{2} T_{1}}$

(B) $\frac{P_{1} V_{1} T_{1}+P_{2} V_{2} T_{2}}{P_{1} V_{1}+P_{2} V_{2}}$

(C) $\frac{P_{1} V_{1} T_{2}+P_{2} V_{2} T_{1}}{P_{1} V_{1}+P_{2} V_{2}}$

(D) $\frac{T_{1} T_{2}\left(P_{1} V_{1}+P_{2} V_{2}\right)}{P_{1} V_{1} T_{1}+P_{2} V_{2} T_{2}}$

JEE Main · 2008 · 27 Apr 2008 · Offline · Heat & Thermodynamics

An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume

V_{1}

and contains ideal gas at pressure

P_{1}

and temperature

T_{1}

. The other chamber has volume

V_{2}

and contains ideal gas at pressure

P_{2}

and temperature

T_{2}

. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be

(A) $\frac{T _{1} T _{2} ( P _{1} V _{1} + P _{2} V _{2} )}{P _{1} V _{1} T _{2} + P _{2} V _{2} T _{1}}$ ✓ Correct
(B) $\frac{P _{1} V _{1} T _{1} + P _{2} V _{2} T _{2}}{P _{1} V _{1} + P _{2} V _{2}}$
(C) $\frac{P _{1} V _{1} T _{2} + P _{2} V _{2} T _{1}}{P _{1} V _{1} + P _{2} V _{2}}$
(D) $\frac{T _{1} T _{2} ( P _{1} V _{1} + P _{2} V _{2} )}{P _{1} V _{1} T _{1} + P _{2} V _{2} T _{2}}$

Answer: (A)

Solution

The container is insulated and the partition is removed without any work being done, so no heat is exchanged and no work is performed; the total internal energy of the gas is conserved. For an ideal gas the internal energy is

n C_{v} T

, and the mole content of each chamber follows from

P V = n R T

, giving

n_{1} = \frac{P _{1} V _{1}}{R T _{1}}

and

n_{2} = \frac{P _{2} V _{2}}{R T _{2}}

. Energy conservation reads

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

; the factor

C_{v}

cancels, and substituting the mole expressions gives

\frac{P _{1} V _{1}}{T _{1}} T_{1} + \frac{P _{2} V _{2}}{T _{2}} T_{2} = (\frac{P _{1} V _{1}}{T _{1}} + \frac{P _{2} V _{2}}{T _{2}}) T

. Solving for the final temperature,

T = \frac{T _{1} T _{2} ( P _{1} V _{1} + P _{2} V _{2} )}{P _{1} V _{1} T _{2} + P _{2} V _{2} T _{1}}

JEE Main · 2008 · 27 Apr 2008 · Offline · Single correct option

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JEE Main · 2008 · 27 Apr 2008 · Offline · Heat & Thermodynamics

Solution

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