Kinds of Thermodynamic Processes

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    Physic

    Kinds of Thermodynamic Processes

    Members of group:

    1. Falah Ibrahim Yazidulhaq (12)2. Fikri Azmi (15)3. Imam Pratama Istiaji (19)4. Noor Taufik (26)5. Yusuf Dewantoro (32)

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    Kinds of Thermodynamic Processes

    I. Intoduction

    In this section, we describe four specific kinds of thermodynamic processes that often

    occur in practical situations. These can be summarized briefly as no heat transfer or adiabatic,

    constant volume or isochoric, constant pressure or isobaric, and constant temperature or

    isothermal. For some of these, we can use a simplified form of the first law of thermodynamics.

    II. Aims

    From this section, we have some aim write these letters. Such as : (1) To solve a Physic Task

    from Mr. Nardi. (2) To know more about thermodynamic process.

    III. Contents Isothermal process ( themperature constant)

    If themperature constant. So,

    In this process T = 0, so the internal energy is also constant, andwork:

    In some special cases, the internal energy of a system depends only on its temperature and not on

    its pressure or volume. The most familiar system having this special property is an ideal gas,

    which we will discuss in the next section. For such systems, if the temperature is constant, theinternal energy is also constant; U=0 and Q=W. That is, any energy entering the system as heat

    Q must leave it again as work W done by the system.

    Isobaric process (pressure constant)

    pV = konstan

    piVi = pfVf

    W = p dVp = nRT/V, so

    W = nRT (1/V) dVW = nRT ln (Vf/Vi)

    Q = W

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    An isobaric process is a constant-pressure process. In general, none of the three quantities

    U, Q and dW is zero in an isobaric process, but calculating W is easy nonetheless.

    If themperature constant. So,

    In this process W = p V = p (Vf- Vi ), so:

    Isochoric process (volume constant)If themperature constant. So,

    In this process V = 0, so W = 0, then:In an isochoric process, all the energy added as heat remains

    in the system as an increase in internal energy. Heating a gas ina closed constant-volume container is an example of an

    isochoric process. (Note that there are types of work that do not

    involve a volume change. For example, we can do work on a

    fluid by stirring it. In some cases, isochoric is used to mean that no work is done).

    Adiabatic process (heat constant)An adiabatic process is defined as one with no heat transfer into or out of a system: Q=0.

    We can prevent heat flow either by surrounding the system with thermally insulating material or

    by carrying out the process so quickly that there is not enough time for appreciable heat flow.

    From the first law, we find that for every adiabatic process,

    If heat constant. So,

    V/T = constant

    Vi/ Ti = Vf/Tf

    U = Q WU = n cpT - p V

    p/T= constantpi/ Ti = pf/Tf

    Q = U = n cvT

    piVi = pfV

    f

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    When a system expands adiabatically. W is positive(the system does work on its surroundings), so U isnegative and the internal energy decreases. When a system

    is compressed adiabatically, W is negative (work is doneon the system by its surroundings) and Uincreases. In

    many (but not all) systems, an increase of internal energy is accompanied by a rise in

    temperature.

    Problems for Practice Problem 1

    The volume of a gas in a container expanded from 1L to 3L upon releasing the piston

    upward. From following graph, find the amount work associated with the expansion of the

    gas!

    from [1, 3]

    Problem 2Calvin is observing an unknown monatomic gas molecule (sealed inside a container) in

    his Freshman Chemistry Lab. He has been told by his lab instructor that there are four moles

    of this unidentified gas in the container. The laboratory rooms temperature was initially setat room temperature when he started the lab, but by the time he was almost finished with the

    lab the temperature has gone up 10C. What is the total internal energy of this unknown

    gaseous substance by the time Calvins lab session ended?

    W = p dVp = k/V

    , k = constant , so:

    W = (k/V) dVW = 1/(1-) { pfVf- piVi} U = -W

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    Solutions for problems Solution of problem 1

    Here, we have a shape that is similar to a rectangle .

    Work = Area

    = Length x Width

    = (3)(3-1) =(3)(2) =(3)(2) =6

    Work = 6 Joules

    Solution of problem 2From the problem, we know that the unknown gaseous substance is monatomic, so we

    would have to use the equation for a monatomic gas:

    Also from the problem, we know that the substance is at room temperature with is 25C,

    but that wasjust extra information given and we didnt really need to know that to solvethe problem. Since we know that the temperature change from start to finish was +10.

    The problem also given us the number of moles that was in the container:

    For the R-value, we can choose any R constant, but to make this problem a little easier

    we will choose a R-constant that will cancel out any other units except for joules.

    Plugging in all these values into the equation we have:

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    IV. Conclusions

    ???

    V. References

    1. http://chemwiki.ucdavis.edu/Wikitexts/UCD_Chem_2BH/ChemWiki_Module_Topics/Ideal_Gas_Processes

    2. http://www.transtutors.com/physics-homework-help/thermal-physics/thermodynamic-process.aspx

    3. http://www.tutorvista.com/content/physics/physics-iii/heat-and-thermodynamics/thermodynamic-processes.php

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