A piston of mass M is hung from a massless spring whose restoring force law goes as F=-k x^3 , where k is the spring constant of appropriate dimension. The piston separates the vertical chamber into two parts, where the bottom part is filled with ' n ' moles of an ideal gas. An external work is done on the gas isothermally (at a constant temperature T) with the help of a heating filament (with negligible volume) mounted in lower part of the chamber, so that the piston goes up from a height L_0 to L_1 , the total energy delivered by the filament is:(Assume spring to be in its natural length before heating)

n R T \ln (\fracL_1L_0)+M g(L_1-L_0)+\frack4(L_1^4-L_0 ^4)

n R T \ln (\fracL_1L_0)+M g(L_1-L_0)+\frac3 k4(L_1 ^4-L_0 ^4)

n R T \ln (\fracL_1^2L_0^2)+\fracM g2(L_1-L_0)+\frack4(L_1^4-L_0 ^4)

3 n R T \ln (\fracL_1L_0)+2 M g(L_1-L_0)+\frack3(L_1 ^3-L_0 ^3)

Heat and Thermodynamics PYQ — JEE Main 2025 · 3 Apr · Shift 1 · Q62 with Solution

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Correct answer: B

Set up the thermodynamics

The gas is heated isothermally at temperature $T$ . For an ideal gas, internal energy depends only on temperature, so for an isothermal process:

\Delta U = 0

Hence, by the first law:

Q = W

where $Q$ is the heat supplied by the filament and $W$ is the total work done by the gas.

So we only need to calculate the work done by the gas in lifting the piston against:

atmospheric/gas pressure effect already included through $PV$ relation,
weight of piston,
spring force.

Geometry and volume change

Let the cross-sectional area of the chamber be $A$ . If piston height from the bottom is $L$ , then gas volume is

V = AL

Thus,

dV = A\,dL

Forces on the piston

At any intermediate height $L$ , the piston is in quasistatic equilibrium.

Upward force by gas:

F_{\text{gas}} = PA

Downward forces:

Weight of piston: $Mg$
Spring restoring force.

Given restoring force law:

F_s = -k x^3

So its magnitude is $k x^3$ , opposite to displacement.

Since the spring is at natural length initially, when piston rises to height $L$ , the extension of spring is proportional to $L$ . Taking the natural reference consistently, the spring contribution to work from $L_0$ to $L_1$ is:

W_s = \int_{L_0}^{L_1} kL^3\,dL = \frac{k}{4}(L_1^4-L_0^4)

Also, work done against gravity is:

W_g = Mg(L_1-L_0)

Work done by gas on itself during isothermal expansion/compression relation

For an ideal gas in an isothermal quasistatic change:

W_{PV} = \int P\,dV = nRT\ln\left(\frac{V_1}{V_0}\right)

Since $V=AL$ ,

\frac{V_1}{V_0} = \frac{AL_1}{AL_0} = \frac{L_1}{L_0}

Therefore,

W_{PV} = nRT\ln\left(\frac{L_1}{L_0}\right)

Total heat supplied by filament

Because the process is isothermal and $\Delta U=0$ ,

Q = W_{\text{total}}

The total energy supplied must account for:

isothermal ideal-gas work term,
gain in gravitational potential energy of piston,
energy stored in spring.

Thus,

Q = nRT\ln\left(\frac{L_1}{L_0}\right)+Mg(L_1-L_0)+\frac{k}{4}(L_1^4-L_0^4)

Match with options

This matches Option B:

nRT\ln \left(\frac{L_1}{L_0}\right)+M g\left(L_1-L_0\right)+\frac{k}{4}\left(L_1^4-L_0^4\right)

Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.

More from Heat and Thermodynamics

View written solutionFree

Correct answer: B

Set up the thermodynamics

The gas is heated isothermally at temperature $T$ . For an ideal gas, internal energy depends only on temperature, so for an isothermal process:

\Delta U = 0

Hence, by the first law:

Q = W

where $Q$ is the heat supplied by the filament and $W$ is the total work done by the gas.

So we only need to calculate the work done by the gas in lifting the piston against:

atmospheric/gas pressure effect already included through $PV$ relation,
weight of piston,
spring force.

Geometry and volume change

Let the cross-sectional area of the chamber be $A$ . If piston height from the bottom is $L$ , then gas volume is

V = AL

Thus,

dV = A\,dL

Forces on the piston

At any intermediate height $L$ , the piston is in quasistatic equilibrium.

Upward force by gas:

F_{\text{gas}} = PA

Downward forces:

Weight of piston: $Mg$
Spring restoring force.

Given restoring force law:

F_s = -k x^3

So its magnitude is $k x^3$ , opposite to displacement.

W_s = \int_{L_0}^{L_1} kL^3\,dL = \frac{k}{4}(L_1^4-L_0^4)

Also, work done against gravity is:

W_g = Mg(L_1-L_0)

Work done by gas on itself during isothermal expansion/compression relation

For an ideal gas in an isothermal quasistatic change:

W_{PV} = \int P\,dV = nRT\ln\left(\frac{V_1}{V_0}\right)

Since $V=AL$ ,

\frac{V_1}{V_0} = \frac{AL_1}{AL_0} = \frac{L_1}{L_0}

Therefore,

W_{PV} = nRT\ln\left(\frac{L_1}{L_0}\right)

Total heat supplied by filament

Because the process is isothermal and $\Delta U=0$ ,

Q = W_{\text{total}}

The total energy supplied must account for:

isothermal ideal-gas work term,
gain in gravitational potential energy of piston,
energy stored in spring.

Thus,

Q = nRT\ln\left(\frac{L_1}{L_0}\right)+Mg(L_1-L_0)+\frac{k}{4}(L_1^4-L_0^4)

Match with options

This matches Option B:

nRT\ln \left(\frac{L_1}{L_0}\right)+M g\left(L_1-L_0\right)+\frac{k}{4}\left(L_1^4-L_0^4\right)

Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.

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