Question

On a warm summer day, a large mass of air (atmospheric pressure$\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{Pa}$) is heated by the ground to$\mathbf{26}\mathbf{°}\mathbf{C}$ 26.0°C and then begins to rise through the cooler surrounding air. (This can be treated approximately as an adiabatic process; why?) Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only$\mathbf{0}\mathbf{.}\mathbf{850}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{Pa}$. Assume that air is an ideal gas, with y= 1.40. (This rate of cooling for dry, rising air, corresponding to roughly 1 C° per 100 m of altitude, is called the dry adiabatic lapse rate.)

Short answer

The temperature of the air mass when it has risen is${\mathrm{T}}_2=284.6\mathrm{K}$ and${\mathrm{T}}_2=11.6\mathrm{C}°$ (In Celsius)

Step-by-step solution

Step 1: Adiabatic ideal gas relation.

An adiabatic process in which no heat transfer takes place between a system and its surroundings. Zero heat transfer is an idealization.

Where,= The ratio of heat capacities$\left[\mathrm{Y}=\frac{{\mathrm{C}}_{\mathrm{P}}}{{\mathrm{C}}_{\mathrm{V}}}\right]$

Thus for an initial state$\left({\mathrm{T}}_1,{\mathrm{V}}_1\right)$and a final state$\left({\mathrm{T}}_2,{\mathrm{V}}_2\right)$

${\mathrm{T}}_1{\left({\mathrm{V}}_1\right)}^{\mathrm{\gamma }-1}={\mathrm{T}}_2{\left({\mathrm{V}}_2\right)}^{\mathrm{\gamma }-1}$(adiabatic process, ideal gas)

${\left(\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}\right)}^{\mathrm{\gamma }-1}={\mathrm{T}}_2{\left({\mathrm{V}}_2\right)}^{\mathrm{\gamma }-1}$………(1)

From the ideal gas equation,

PV = nRT

Where n = Number of moles , R= universal gas constant , V = Volume of gas ,

p = Pressure, and T = Temperature of gas

From initial state$\left({\mathrm{p}}_1,{\mathrm{T}}_1,{\mathrm{V}}_1\right)$and the final state$\left({\mathrm{p}}_2,{\mathrm{T}}_2,{\mathrm{V}}_2\right)$relation of ideal gas equation becomes,

$$\begin{gathered} \frac{{\mathrm{p}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{p}}_2{\mathrm{V}}_2}{{\mathrm{T}}_2} \\ \frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}=\frac{{\mathrm{T}}_1{\mathrm{p}}_2}{{\mathrm{T}}_2{\mathrm{p}}_1} \end{gathered}$$

Substitute the value of$\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}$ in equation 1^stWe get

${\left(\frac{{\mathrm{T}}_1{\mathrm{p}}_2}{{\mathrm{T}}_2{\mathrm{p}}_1}\right)}^{\mathrm{\gamma }-1}=\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}$…..(2)

Calculate the temperature of final state

We are given a large mass of air at the atmospheric pressure and let us consider this pressure is the initial pressure ${\mathrm{p}}_1$where role="math" localid="1664297754243" ${\mathrm{p}}_1=1.01\times {10}^5\mathrm{pa}$at initial temperature T1 : 26.0 C° (299 K). This air rise through the cooler surrounding air and this can be treated as an adiabatic process because the hot air has a work done against the gravity and expands up to high and we want to calculate the final temperature T2 of the air after raising when the pressure becomes ${\mathrm{p}}_2=0.850\times {10}^2\mathrm{pa}$where the air is considered to be an ideal gas with y = 1.40

From equation 2^nd

^{role="math" localid="1664298058353" $$\begin{gathered} {\left(\frac{{\mathrm{T}}_1}{\mathrm{T}2}\right)}^{\mathrm{\gamma }-1}={\left(\frac{{\mathrm{p}}_2}{{\mathrm{p}}_1}\right)}^{\mathrm{\gamma }-1}=\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1} \\ {\left(\frac{{\mathrm{p}}_2}{{\mathrm{p}}_1}\right)}^{\mathrm{\gamma }-1}=\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\times {\left(\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\right)}^{\mathrm{y}-1} \\ {\left(\frac{{\mathrm{p}}_2}{{\mathrm{p}}_1}\right)}^{\mathrm{\gamma }-1}={\left(\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\right)}^{\mathrm{y}} \\ {\left(\frac{{\mathrm{p}}_2}{{\mathrm{p}}_1}\right)}^{\frac{\mathrm{\gamma }-1}{\mathrm{\gamma }}}=\left(\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}\right) \end{gathered}$$}

Substitute all the given values in equation in above equation

${\mathrm{T}}_2={\mathrm{T}}_1{\left(\frac{{\mathrm{p}}_2}{{\mathrm{p}}_1}\right)}^{\frac{\mathrm{y}-1}{\mathrm{y}}}$

$$\begin{gathered} {\mathrm{T}}_2=299\mathrm{K}\times {\left(\frac{0.850\times {10}^5\mathrm{Pa}}{1.01\times {10}^5\mathrm{Pa}}\right)}^{\frac{\left(1.4-1\right)}{1.4}} \\ {\mathrm{T}}_2=284.6\mathrm{K} \end{gathered}$$

Hence , The temperature of the air mass when it has risen is${\mathrm{T}}_2=284.6\mathrm{K}$ and${\mathrm{T}}_2=11.6C°$ (In Celsius)