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A monatomic ideal gas expands slowly to twice its orig- inal volume, doing 450 J of work in the process. Find the heat added to the gas and the change in internal energy of the gas if the process is
(a) isothermal;
(b) adiabatic;
(c) isobaric.

Respuesta :

Answer:

a)  Q = - 450 J  ΔE= 0 b)  ΔE = 450 J  Q=0

Explanation:

To find the heat transferred in the process let's use the first law of thermodynamics

    ΔE = Q + W

Let's apply this equation every case.

Isothermal process.  This is done at a constant temperature so that internal energy change is zero.

 

    0 = Q + W

    Q = -W

    Q = - 450 J

 

Adiabatic process.  In this process the system is isolated, so there is no heat transfer (Q = 0)

   

    ΔE = W

    ΔE = 450 J

    Q = 0

Isobaric process.  Work is done by changing the volume at constant pressure.  In this case the internal energy and heat changes in the process, so we will calculate with the equation

     W = - P ΔV

     P = W ΔV

     P = 450 / (2V-V)

     P = 450 / V      Pa

From here we can find the temperature using the ideal gas equation

   

     PV = n R T

Let's write this equation for the two volumes, assuming the constant pressure

     P V1 = mR T1

     P V2 = n R T2

     P (v2-V1) = n R (T2-T1)

     450 = nR (T2-T1)

     ΔT = 450 /nR

We can use the Boltzmann equation that relates energy and temperature

     E = KT

     ΔE = K ΔT

     ΔE = K 450 / n R

Boltzmann's constant is

     K = R / n

replacing

      ΔE = (R / n) 450 / nR

      ΔE = 450 n²

We already have the two terms of the first law of thermodynamics

   Q = DE - W

   Q = 450 / n² - 450

 Q = 450 (1/n²  -1)

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