Question

The rectangular box shown in Fig has a partition which can slide without friction along the length of the box. Initially, each of the two chambers of the box has one mole of a mono-atomic ideal gas (γ=5/3) at a pressure P0, volume V0 and temperature T0. The chamber on the left is slowly heated by an electric heater. The walls of the box and the partition are thermally insulated. Heat loss through the lead wires of the heater is negligible. The gas in the left chamber expands pushing the partition, until the final pressure in both chambers becomes 243P0/32. Determine the work done by the gas in the right chamber.

[Take R=8.3 J/K mol]

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Solution

The correct option is **C** −15.5 T0

For the left chamber,

P0V0T0=P0×24332×T1×V1

(nR=constant)

⇒T1=24332×V1T0V0 ......(1)

For the right chamber for adiabatic compression, we get,

P0Vγ0=P0×24332×Vγ2

⇒V2V0=(32243)3/5⇒V2=827V0

But V1+V2=2V0

∴V1=2V0−V2=2V0−827V0

⇒V1=4627V0 .........(2)

From (1) and (2),

T1=24332×46×V0V0×27×T0

or T1=20716T0=12.9 T0 (approx.)

To find the temperature in the second chamber (right), we apply

(T1T2)γ=(P2P1)1−γ

⇒(T0T2)5/3=(243P032P0)1−5/3

⇒T2=2.25 T0

Work done in right chamber (adiabatic process)

W=11−γ(P2V2−P0V0)

=−32[(24332P0×827V0)−P0V0]

=−32(94−1)P0V0=−158×RT0

⇒W=−15.5 T0

For the left chamber,

P0V0T0=P0×24332×T1×V1

(nR=constant)

⇒T1=24332×V1T0V0 ......(1)

For the right chamber for adiabatic compression, we get,

P0Vγ0=P0×24332×Vγ2

⇒V2V0=(32243)3/5⇒V2=827V0

But V1+V2=2V0

∴V1=2V0−V2=2V0−827V0

⇒V1=4627V0 .........(2)

From (1) and (2),

T1=24332×46×V0V0×27×T0

or T1=20716T0=12.9 T0 (approx.)

To find the temperature in the second chamber (right), we apply

(T1T2)γ=(P2P1)1−γ

⇒(T0T2)5/3=(243P032P0)1−5/3

⇒T2=2.25 T0

Work done in right chamber (adiabatic process)

W=11−γ(P2V2−P0V0)

=−32[(24332P0×827V0)−P0V0]

=−32(94−1)P0V0=−158×RT0

⇒W=−15.5 T0

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