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Question

An ideal gas undergoes a cyclic process abcda which is shown by a pressure - density curve. Identify the option with incorrect statement.


A
Work done by the gas in process bc is zero
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B
Work done by the gas in process cd is negative
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C
Internal energy of the gas at state a is greater than that at state c
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D
Net work done by the gas in the cycle is negative
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Solution

The correct option is C Internal energy of the gas at state a is greater than that at state c
From the ideal gas equation,
PM=ρRT
Pρ=RTM
Here, RTM is the slope of Pρ curve.
As R and M are constant for a gas,
Slope (m)T
From graph, mcd>mab, thus, Tc>Ta
Also, the slope of the processes ab and cd are constant, thus, temperature is constant in these processes.
ab and cd are isothermal processes.
For a given gas, ρ=MV.
For processes bc and da, density is constant, thus, volume will be constant.
bc and da are isochoric processes.
Plotting the PV diagram, we get,


For option (a),
In process bc, as there is no change in volume, the work done by gas is zero.
For option (b),
In process cd, as volume is decreasing and pressure is increasing, thus it is a compression process. Therefore work done by gas is negative.
For option (c),
As Tc>Ta, internal energy at c is greater than at a.
For option (d),
As the process is cyclic anticlockwise, thus, the work done by gas is negative.
Hence, option (c) contains incorrect statement.

Why this question? It provides unique challenge to apply laws of thermodynamics and ideal gas equation. Caution: For comparing work done and change in internal energy in a given process, always convert given curve to P - V diagram.

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