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Which one of the following statements is false? [2001S] (a) Work is a state function. (b) Temperature is a state function. (c) Change in the state is completely defined when the initial and final states are specified. (d) Work appears at the boundary of the system.

Short Answer

Expert verified
Statement (a) is false because work is not a state function.

Step by step solution

01

Understanding State Function

A state function is a property of a system that depends only on the initial and final states and not on the path taken. Temperature is a classic example, as it is independent of the process path.
02

Evaluating Each Statement

Let's evaluate each statement: - (a) Work is a state function: This is false because work depends on the path taken. - (b) Temperature is a state function: This is true as temperature only depends on the state of the system. - (c) Change in the state is completely defined when the initial and final states are specified: This is true for state functions. - (d) Work appears at the boundary of the system: This is true as work is one of the forms of energy transfer across the boundary.
03

Identifying the False Statement

From the evaluations: - Statement (a) is false because work is not a state function, it is a path function. Other statements (b), (c), and (d) are true. Thus, the false statement is statement (a).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

State Function
In the realm of thermodynamics, a state function is crucial because it informs us about the property of a system based solely on its current state, irrespective of the journey it took to get there. This means that if you know the initial and final states of a system, you can determine the change in the state function, without concerning yourself with the process path. Examples of state functions include:
  • Temperature
  • Pressure
  • Volume
  • Internal energy
No matter how the system changes from one state to another, whether rapidly or slowly, the change in a state function remains constant. This is because state functions are independent of the process or path. So, when we only care about where a system starts and ends, state functions become particularly useful. They provide simplified calculations in thermodynamic equations by allowing us to focus on the initial and final states alone.
Path Function
Unlike state functions, path functions are properties that depend on the specific transitions or paths between states. In simpler terms, they are influenced by HOW the system goes from one state to another. Work and heat are classic examples of path functions in thermodynamics. To further break this down:
  • Work: It can vary significantly based on the method, route, or pathway taken to implement a process.
  • Heat: Similarly, the amount of heat transferred depends on the exact process undergone.
The dependency on paths means that integrating path functions over different paths yields different results, as opposed to state functions which only care about the end states. Thus, work done during a thermodynamic process cannot be determined solely by knowing the initial and final states, but instead requires detailed path data.
Work in Thermodynamics
Work, in thermodynamics, refers to the energy transfer that occurs when a force is applied over a distance or against a pressure. It is the form of energy that crosses the boundary of a system and can alter the system's state. Here are some important points about work in thermodynamics:
  • Work is a path function, meaning it depends on the specific sequence of events leading between two states.
  • Work appears and disappears across the boundary, often visible as changes in the surroundings, like stirring or compressing gases.
  • Since it relies on the process, calculations of work can vary depending on compressing methods or expansion for example.
The unit of work is the joule (J), and its calculation often involves the integration of pressure-volume curves or force-distance scenarios. Understanding work as a path function, versus the simpler calculations of a state function, underscores why thermodynamics often focuses heavily on detailed path-dependent evaluations.

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Most popular questions from this chapter

Read the following statement and explanation and answer as per the options given below : Assertion : The heat absorbed during the isothermal expansion of an ideal gas against vacuum is zero. Reason : The volume occupied by the molecules of an ideal gas is zero. (a) If both assertion and reason are CORRECT, and reason is the CORRECT explanation of the assertion. (b) If both assertion and reason are CORRECT, but reason is NOT the CORRECT explanation of the assertion. (c) If assertion is CORRECT, but reason is INCORRECT. (d) If assertion is INCORRECT, but reason is CORRECT.

In order to get maximum calorific output, a burner should have an optimum fuel to oxygen ratio which corresponds to 3 times as much oxygen as is required theoretically for complete combustion of the fuel. A burner which has been adjusted for methane as fuel (with \(x\) litre/hour of \(\mathrm{CH}_{4}\) and \(6 x\) litre/hour of \(\mathrm{O}_{2}\) ) is to be readjusted for butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\). In order to get the same calorific output, what should be the rate of supply of butane and oxygen ? Assume that losses due to incomplete combustion, \(e t c\), are the same for both the fuels and the gases behave ideally. Heats of combustion : $$ \mathrm{CH}_{4}=809 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{C}_{4} \mathrm{H}_{10}=2878 \mathrm{~kJ} / \mathrm{mol} $$

The standard molar enthalpies of formation of cyclohexane(1)and benzene(1) at \(25^{\circ} \mathrm{C}\) are \(-156\) and \(+49 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. The standard enthalpy of hydrogenation of cyclohexene(1) at \(25^{\circ} \mathrm{C}\) is \(-119\) \(\mathrm{kJ} \mathrm{mol}^{-1}\). Use these data to estimate the magnitude of the resonance energy of benzene.

In thermodynamics, a process is called reversible when (a) surroundings and system change into each other. (b) there is no boundary between system and surroundings. (c) the surroundings are always in equilibrium with the system. (d) the system changes into the surroundings spontaneously.

The standard state Gibbs free energies of formation of C(graphite) and \(\mathrm{C}\) (diamond) at \(T=298 \mathrm{~K}\) are \(\Delta_{f} G^{0}[\mathrm{C}\) (graphite) \(]=0 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\Delta_{f} G^{0}[\mathrm{C}\) (diamond) \(]=2.9 \mathrm{~kJ} \mathrm{~mol}^{-1}\) The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite \([\mathrm{C}\) (graphite) \(]\) to diamond \(\left[C\right.\) (diamond)] reduces its volume by \(2 \times 10^{-}\) \({ }^{6} \mathrm{~m}^{3} \mathrm{~mol}^{-1} .\) If \(C\) (graphite) is converted to \(\mathrm{C}\) (diamond) isothermally at \(T\) \(=298 \mathrm{~K}\), the pressure at which \(\mathrm{C}\) (graphite) is in equilibrium with \(\mathrm{C}\) (diamond), is [Useful information : \(\left.1 \mathrm{~J}=1 \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2} ; 1 \mathrm{~Pa}=1 \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-2} ; 1 \mathrm{bar}=10^{5} \mathrm{~Pa}\right]\) [Adv. 2017] (a) 14501 bar (b) 58001 bar (c) \(1450 \mathrm{bar}\) (d) 29001 bar

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