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Chapter 8: Problem 1104

If du represents the increase in internal energy of a thermodynamic system and dw the work done by the system, which of the following statement is true? (A) \(\mathrm{du}=\mathrm{dw}\) in isothermal process (C) \(\mathrm{du}=-\mathrm{dw}\) in an aidabatic process (B) \(\mathrm{du}=\mathrm{dw}\) in aidabatic process (D) \(\mathrm{du}=-\mathrm{dw}\) in an isothermal process

Short Answer

Expert verified
The correct statement is (C): \(\mathrm{du}=-\mathrm{dw}\) in an adiabatic process.

Step by step solution

01

Check each statement based on thermodynamics understanding

Firstly, understanding the nature of each process will guide through the evaluation of each statement.
02

Evaluate statement (A)

Statement A, " \(\mathrm{du}=\mathrm{dw}\) in an isothermal process", is incorrect because in an isothermal process the internal energy (\(\mathrm{du}\)) doesn't change, the work done by the system isn't equal to the change in internal energy but is equal to the heat transferred.
03

Evaluate statement (C)

Statement C, "\(\mathrm{du}=-\mathrm{dw}\) in an adiabatic process", is true. In an adiabatic process, there is no heat transfer (\(\mathrm{dq}=0\)), so the change in internal energy (\(\mathrm{du}\)) is equal to the negative of the work done by the system, according to the first law of thermodynamics.
04

Evaluate statement (B)

Statement B, "\(\mathrm{du}=\mathrm{dw}\) in an adiabatic process", is incorrect. This is because, similar to statement C, in an adiabatic process the change in internal energy (\(\mathrm{du}\)) is equal to the negative work done by the system and not equal to the work done by the system.
05

Evaluate statement (D)

Lastly, statement D, "\(\mathrm{du}=-\mathrm{dw}\) in an isothermal process", is incorrect. During an isothermal process, the internal energy does not change, meaning \(\mathrm{du}=0\), not \(-\mathrm{dw}\). In conclusion, the correct statement among the given options is statement C: "\(\mathrm{du}=-\mathrm{dw}\) in an adiabatic process".

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Most popular questions from this chapter

In a thermodynamic process, pressure of a fixed mass of a gas is changed in such a manner that the gas release \(20 \mathrm{~J}\) of heat and $8 \mathrm{~J}$ of work has done on the gas-If the initial internal energy of the gas was \(30 \mathrm{j}\), then the final internal energy will be (A) \(58 \mathrm{~J}\) (B) \(2 \mathrm{~J}\) (C) \(42 \mathrm{~J}\) (D) \(18 \mathrm{~J}\)

If a heat engine absorbs \(50 \mathrm{KJ}\) heat from a heat source and has efficiency of \(40 \%\), then the heat released by it in heat sink is (A) \(40 \mathrm{KJ}\) (B) \(30 \mathrm{KJ}\) (C) \(20 \mathrm{~J}\) (D) \(20 \mathrm{KJ}\)

A Carnot engine Converts one sixth of the heat input into work. When the temperature of the sink is reduced by \(62^{\circ} \mathrm{C}\) the efficiency of the engine is doubled. The temperature of the source and sink are (A) \(80^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}\) (B) \(95^{\circ} \mathrm{C}, 28^{\circ} \mathrm{C}\) (C) \(90^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}\) (D) \(99^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}\)

For an adiabatic process involving an ideal gas (A) \(\mathrm{P}^{\gamma-1}=\mathrm{T}^{\gamma-1}=\) constant (B) \(\mathrm{P}^{1-\gamma}=\mathrm{T}^{\gamma}=\) constant (C) \(\mathrm{PT}^{\gamma-1}=\) constant (D) \(\mathrm{P}^{\gamma-1}=\mathrm{T}^{\gamma}=\) constant

For an isothermal expansion of a Perfect gas, the value of $(\Delta \mathrm{P} / \mathrm{P})$ is equal to (A) \(-\gamma^{(1 / 2)}\\{(\Delta \mathrm{V}) / \mathrm{V}\\}\) (B) \(-\gamma\\{(\Delta \mathrm{V}) / \mathrm{V}\\}\) (C) \(-\gamma^{2}\\{(\Delta \mathrm{V}) / \mathrm{V}\\}\) \((\mathrm{D})-(\Delta \mathrm{V} / \mathrm{V})\)
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