Question

A uniform cylinder of length l has a thin
movable piston at its middle. Both part of
the tube contains ideal gas at pressure P
and temperature T. When one part is
heated to 2T and other half part is cooled
to T/2, the piston moves by a distance of
(A)
3l/10
(B)
3l/5
(C)
5l/7
(D)
l/10​

Answer 1

When one part is  heated to 2T and other half part is cooled  to T/2, the piston moves by a distance of 3l/10

Therefore, option (A) is correct.

Explanation:

Let the area of cross section of the cylinder be $A$

Given

Length of the cylinder = $l$

Initially since both the gases are at same pressure and temperature and also have the same volume

Initial Volume of gases $V=A\frac{l}{2}$

Therefore, the amount of gases is also the same

For initial condition for both the gases

$PV = nRT$

$\implies \frac{PV}{T}=\text{ constant}$

When one gas is heated to $2T$, it will expand and the other gas is cooled by $T/2$, it will contract

If change in volume is $\Delta V$

The pressure on both side of the piston will be the same if the system reaches equilibrium inside the cylinder

If the new pressure of the gases after they reach equilibrium again is $P'$

Then

For 1st gas

$\frac{PV}{T}=\frac{P'(V+\Delta V)}{2T}$

$\implies 2PV=P'(V+\Delta V)$   ...... (1)

For 2nd gas

$\frac{PV}{T}=\frac{P'(V-\Delta V)}{T/2}$

$\implies PV/2=P'(V-\Delta V)$   ......... (2)

From (1) and (2)

$4=\frac{V+\Delta V}{V-\Delta V}$

$\implies 4V-4\Delta V=V+\Delta V$

$\implies 3V=5\Delta V$

Initial Volume $V=\frac{Al}{2}$

If the piston moves by $x$

Then $\Delta V=Ax$

Thus,

$3A\frac{l}{2}=5Ax$

$\implies x=\frac{3l}{10}$

Hope this answer is helpful.

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