Collection of Solved Problems in Physics

Temperature and work of air

Task number: 2118

• 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.

• Adiabatic (Poisson) equation

  The equation applies to reversible adiabatic processes of an ideal gas and reads:

  \[pV^{\kappa}=konst,\]

  where p is pressure, V is volume and κ is the Poisson constant.

  But we need to adjust this relationship to a form in which the temperature is present.

  For the air that we consider to be an ideal gas, the general gas equation also applies

  \[\frac{pV}{T}=const.\]

  Let us express the pressure p fom the gas equation and substitute it into the adiabatic equation to get the desired dependency.

• Adiabatic (Poisson) equation with temperature

  We express the pressure p from the general gas equation:

  \[\frac{pV}{T}=\mbox{const.} \qquad\Rightarrow \qquad p=\frac{T}{V}\,\cdot\,\mathrm{const.},\]

  and we put it into the adiabatic equation and get

  \[\frac{T}{V}V^{\kappa}=\mathrm{const.}\]

  We have moved the constants to the right-hand side and put them in a single constant. A simple rearrangement yields the searched relationship in the form:

  \[TV^{\kappa-1}=\mathrm{const.}\]

• 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 V_í and V_f 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)

• Boyle-Mariotte law

  According to Boyle-Mariotte law, it holds that:

  \[ pV = const,\]

  where p is the pressure and V is the volume of the gas during an isothermal process

• 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 V_i and V_f 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:

  \[W= p_1V_1\left[\ln{\frac{1}{5}}+\frac{1}{\kappa - 1}\left(1-10^{1-\kappa} \right)\right].\] \[W= 10^5\cdot{10^{-2}}\cdot \left[\ln{\frac{1}{5}}+ \frac{1}{1-1.4}\cdot \left( 1-10^{1-1.4}\right) \right]\,\mathrm{J}\] \[W\dot{=}-3\,114\,\mathrm{J}\]

• Answer

  The resulting temperature is approximately 109 K.

  The gas did work of approximately 3114 J.