Temperature and work of air

Task number: 2118

We have 10 liters of air at a temperature of 273 K and at a pressure of 100 kPa. First, we isothermally compress it to one fifth of its original volume and then we adiabatically expand it to double its original volume. What will be the resulting temperature of the air after the adiabatic expansion and what work will the gas have done throughout the whole process?

Consider the air to be an ideal gas with the heat capacity ratio of κ = 1.4.

  • Notation

    V1 = 10 l = 10−2 m3 Initial volume of air
    T1 = 273 K Initial temperature of air
    p1 = 100 kPa = 105 Pa Initial pressure of air
    V2 = V1/5 Volume of the air after isothermal compression
    V3 = 2V1 Volume of air after adiabatic expansion
    κ = 1.4 Heat capacity ratio
    T3 = ? Temperature after adiabatic expansion
    W = ? Work done throughout whole process
  • Hint 1 – Resulting temperature

    In order to calculate the temperature of the gas after adiabatic expansion, use the adiabatic (Poisson) equation. Modify it so that the temperature appears in it.

  • Analysis – the resulting temperature

    The temperature of the gas does not change during the isothermal compression. Therefore, the gas will have the same temperature before the adiabatic expansion as it had at the beginning of the process.

    We use the adiabatic law in the form in which the gas volume and temperature appear in order to calculate the resulting temperature.

  • Calculation of the resulting temperature

    The temperature of the gas does not change during the isothermal compression. This is described by the equality T1= T2, where T2 is the temperature of the air after compression

    However, things are different in the subsequent expansion.

    For the adiabatic process, there holds the adiabatic (Poisson) law, which we can write in the form

    \[V^{\kappa-1}T=const.\]

    (This relationship can be obtained from the usually used form pV κ = const. by a simple arrangement done in Hint 1)

    In our case, it thus holds that

    \[V_2^{\kappa-1}T_2=V_3^{\kappa-1}T_3.\]

    From this formula we express the resulting temperature T3:

    \[T_3=\left(\frac{V_2}{V_3}\right)^{\kappa-1}T_2\]

    and we substitute the ratio between the volumes V2 and V3 given in the assignment:

    \[T_3=\left(\frac{\frac{1}{5}V_1}{2V_1}\right)^{\kappa-1}T_2= 0.1^{\kappa-1}T_2. \]
  • Hint 2 – calculation of work W

    We will determine the total work W made by the gas as the sum of the work W1 made during the isothermal compression and the work W2 done during the adiabatic expansion

    We need to use the integral calculus to compute the work done in individual processes because the gas pressure changes (it is a function of volume). It holds that

    \[W = \int\limits_{V_i}^{V_f}p\left(V\right)\, \text{d}V,\]

    where Ví and Vf are the initial and the final folume of the gas, respectively.

  • Hint 3 – expressing the pressure p(V)

    To express the volume-dependent pressure p(V) in the case of the isothermal process, we use the Boyle-Mariotte law.

    We express the pressure p(V) using the adiabatic (Poisson) law in the case of the adiabatic process (see Hint 1)

  • Analysis – the total work

    The total work done by the gas is equal to the sum of the work done during isothermal compression and the work done during adiabatic expansion.

    We need to use the integral calculus for both processes, because the pressure is a function of volume in both cases.

    To express the pressure as a function of volume we use the Boyle-Mariott law during the isothermal process and the adiabatic (Poisson) equatin during the adiabatic process. We integrate the acquired functions with respect to volume. We use the initial and the final gas volume as the limits.

  • Calculation of the work done

    The total work W made by the gas is the sum of the work W1 made during the isothermal compression and the work W2 done during the adiabatic expansion.

    For a non-constant pressure p, the general integral relation for work holds

    \[W = \int\limits_{V_i}^{V_f}p\left(V\right)\, \text{d}V,\]

    where Vi and Vf are the initial and the final folume of the gas, respectively.

    In the case of the isothermal process, we can express pressurep as a function of volume V using the Boyle-Mariotte Law

    \[p_1V_1 = pV \qquad \Rightarrow \qquad p = \frac{p_1V_1}{V}.\]

    We substitute into the formula for work

    \[W_1 = \int\limits_{V_1}^{V_2}p\, \text{d}V = \int\limits_{V_1}^{\frac{1}{5}V_1}\frac{p_1V_1}{V}\, \text{d}V =\]

    we factor out the constants in front of the integral

    \[=p_1V_1 \int\limits_{V_1}^{\frac{1}{5}V_1}\frac{1}{V}\, \text{d}V =\]

    we integrate the function and evaluate it at the limits

    \[=p_1V_1[\ln\,V]_{V_1}^{\frac{1}{5}V_1} = p_1V_1\,\ln \,\frac{\frac{1}{5}V_1}{V_1}= p_1V_1\,\ln\,\frac{1}{5}.\]

    In the case of the adiabatic process, we express the pressure p as a function of volume V using the Poisson law

    \[p_2V_{2}^{\kappa} = pV^{\kappa} \qquad \Rightarrow \qquad p = \frac{p_2V_2^{\kappa}}{V^{\kappa}}.\]

    Now we put the obtained expression into the formula for work

    \[W_2 = \int\limits_{V_2}^{V_3}p \, \text{d}V = \int\limits_{V_2}^{V_3}\frac{p_2V_{2}^{\kappa}}{V^{\kappa}} \, \text{d}V =\]

    we factor out the constants in front of the integral

    \[=p_2V_{2}^{\kappa} \int\limits_{V_2}^{V_3}\frac{1}{V^{\kappa}} \, \text{d}V =\]

    we integrate

    \[= p_2V_{2}^{\kappa}\frac{1}{-\kappa + 1}\left[V^{-\kappa + 1}\right]_{V_2}^{V_3}=\]

    we evaluate the integral at the limits

    \[= \frac{p_2V_{2}^{\kappa}}{-\kappa + 1}\left[(V_3)^{-\kappa + 1} - (V_2)^{-\kappa + 1}\right].\]

    We use Boyle-Mariotte law to express the pressure p2 in terms of p 1

    \[p_1V_1=p_2V_2,\]

    from which we get

    \[p_2=\frac{p_1V_1}{V_2}=\frac{p_1V_1}{\frac{1}{5}V_1}=5p_1.\]

    We put this formula, together with the given relations for volumes V2 and V3, into the formula for work

    \[W_2= \frac{5p_1\left(\frac{1}{5}V_{1}\right)^{\kappa}}{1 -\kappa}\left[(2V_1)^{1-\kappa} - \left(\frac{1}{5}V_{1}\right)^{1-\kappa}\right]\]

    and we rearrange the expression

    \[W_2= \frac{p_1V_{1}}{1-\kappa}5^{1-\kappa}\left[2^{1-\kappa} - \left(\frac{1}{5}\right)^{1-\kappa}\right].\]

    Finally, we get

    \[W_2 = \frac{p_1V_{1}}{1-\kappa}\left[10^{1-\kappa} - 1\right].\]

    The total work done by the gas is then

    \[W=W_1+W_2\] \[W=p_1V_1\ln{\frac{1}{5}}+\frac{p_1V_{1}}{1-\kappa}\left(10^{1-\kappa} - 1\right)\] \[W= p_1V_1\left[\ln{\frac{1}{5}}+\frac{1}{\kappa - 1}\left(1-10^{1-\kappa} \right)\right].\]
  • Numerical substitution

    Resulting temperature:

    \[T_3= 0.1^{\kappa-1}T_2 = 0.1^{1.4-1}\cdot 273\,\mathrm{K}\dot{=} 109\,\mathrm{K}\]

    Total work done:

    \[W= p_1V_1\left[\ln{\frac{1}{5}}+\frac{1}{\kappa - 1}\left(1-10^{1-\kappa} \right)\right].\] \[W= 10^5\cdot{10^{-2}}\cdot \left[\ln{\frac{1}{5}}+ \frac{1}{1-1.4}\cdot \left( 1-10^{1-1.4}\right) \right]\,\mathrm{J}\] \[W\dot{=}-105\,\mathrm{J}\]
  • Answer

    The resulting temperature is approximately 109 K.

    The gas did work of approximately 105 J.

Difficulty level: Level 4 – Undergraduate level
Cs translation
Send comment on task by email