Question

An ideal gas undergoes an adiabatic compression from p=1.0atm,$\mathit{V}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathit{L}$,T=0C , p=1.0$\times$10³L and. (a) Is the gas monatomic, diatomic, or polyatomic? (b) What is its final temperature? (c) How many moles of gas are present? What is the total translational kinetic energy per mole (d) before and (e) after the compression? (f) What is the ratio of the squares of the speeds before and after the compression?

Short answer

1. The gas is monoatomic.
2. The final temperature is 2.7*10⁴K.
3. The number of moles are $4.5\times {10}^{4}K$.
4. Total translational kinetic energy per mole before the compression is role="math" localid="1662449550058" $3.4\times {10}^{3}J$
5. Total translational kinetic energy per mole after the compression is $3.4\times {10}^{5}J$.
6. The ratio of the squares of the speeds before and after the compression is 0.010.

Step-by-step solution

Write the given data from the question:

• At Pressure P=1.0atm , volume V=1.0$\times$10⁶ L and temperature T=0^.C
• At Pressure P=1.0$\times$10⁵atm, volume V=1.0$\times$10³L

Understanding the concept

In case of adiabatic process,

$P{V}^{\gamma }=Cons\tan t$

Here P is the pressure, V is the volume and$\gamma$is the ratio specific heat capacity at constant pressure to the specific heat capacity at constant volume.

$\gamma =\frac{C_P}{C_w}$

Here C_P is specific heat capacity at constant pressure and C_Vis specific heat capacity at constant volume.

Work done in case of adiabatic process is given by,

$W=\frac{P_i{V}_i-P_f{V}_f}{Y-1}$

(a) Determine whether the gas is monoatomic or diatomic

For adiabatic compression between two states,

$P_i{V_i}^{\gamma }=P_f{V_f}^{\gamma }$

$\left(\frac{P_i}{P_f}\right)=\frac{V_f^{\gamma }}{V_i^{\gamma }}$

Take both sides

$$\begin{gathered} In\left(\frac{P_i}{P_f}\right)=\gamma In\left(\frac{V_f}{V_i}\right) \\ \gamma =\frac{In\left({\displaystyle \frac{P_i}{P_f}}\right)}{In\left({\displaystyle \frac{V_f}{V_i}}\right)} \end{gathered}$$

Substitute 1.0mol for P_i,$1.0\times {10}^{5}atm$for P_f , $1.0\times {10}^{3}L$for V_fand $1.0\times {10}^{6}L$for V_iinto the above equation,

$$\begin{gathered} \gamma =\frac{In\left({\displaystyle \frac{1.0atm}{1.0\times {10}^{5}atm}}\right)}{In\left({\displaystyle \frac{1.0\times {10}^{3}L}{1.0\times {10}^{6}L}}\right)} \\ =\frac{5}{3} \end{gathered}$$

Therefore the gas is monoatomic.

(b) Determine the final temperature.

Applying the ideal gas equation for initial and final state

The initial state equation,

$P_i{V}_I=nR{T}_i$

The final state equation,

$P_f{V}_f=nR{T}_f$

Therefore, the ratio of above two equations is,

$\frac{P_i{V}_i}{P_f{V}_f}=\frac{T_i}{T_f}$

role="math" localid="1662451334723" $\Rightarrow T_f=T_i\left(\frac{P_f{V}_f}{P_i{V}_i}\right)$

Substitute 273K for T_i , $1.0x{10}^{5}atm$for P_f ,$1.0\times {10}^{3}L$ for V_f , 1.0atm for P_iand $1.0\times {10}^{6}L$for V_I into the above equation,

$T_f=\left(273K\right)\left[\frac{\left(1.0\times {10}^{5}atm\right)\left(1.0\times {10}^{3}L\right)}{\left(1.0atm\right)\left(1.0\times {10}^{6}L\right)}\right]$

$=2.7\times {10}^{4}K$

Therefore, thefinal temperature is 2.7$\times$10⁴K.

(c) Determine the final temperature.

The ideal gas equation is

$P_i{V}_i=nR{T}_i$
$\Rightarrow n=\frac{P_i{V}_i}{R{T}_i}$

Substitute $1.0x{10}^{5}atm$for P_i, $1.0\times {10}^{3}L$for V_i , $8.31\frac{J}{mol}K$for R, 273K for T_iinto the above equation,

$$\begin{gathered} n=\frac{\left(1.0\times {10}^{5}atm\right)\left(1.0\times {10}^{3}L\right)}{\left(8.31{\displaystyle \frac{J}{mol}}K\right)\left(273K\right)} \\ =4.5\times {10}^{4}mol \end{gathered}$$

Therefore, the number of moles is $4.5\times {10}^{4}mol$.

(d) Calculate the total translational kinetic energy before the compression.

The expression for the translational kinetic energy before the compression is,

$K_i=\frac{3}{2}R{T}_i$

Substitute $8.31\frac{J}{mol}K$for R, 273K for T_i into the above equation,

$$\begin{gathered} K_i=\frac{3}{2}\left(8.31\frac{J}{mol}K\right)\left(273K\right) \\ =3.4\times {10}^{3}J \end{gathered}$$

Therefore, translational kinetic energy per mole before the compression is $3.4\times {10}^{3}J$.

(e) Calculate the total translational kinetic energy after the compression.

The expression for the translational kinetic energy before the compression is,

$K_f=3/2R{T}_f$

Substitute $8.31\frac{J}{mol}K$for R, localid="1662454671649" $2.7\times {10}^{4}K$ for T_f into the above equation,

$$\begin{gathered} K_f=\frac{3}{2}\left(8.31\frac{J}{mol}K\right)\left(273K\right) \\ =3.4\times {10}^{5}J \end{gathered}$$

Therefore, the translational kinetic energy per mole after the compression is 3.4$\times$10⁵J.

(f) Calculate the total translational kinetic energy after the compression.

The square of rms velocity is directly proportional to the temperature,

$V_{rms}^2\propto T$

So,

$\frac{V_{rms,i}^2}{V_{rms,f}^2}=\frac{T_i}{T_f}$

Substitute 273K for T_i and $2.7\times {10}^{4}K$for T_f into the above equation,

$=\frac{\left(273K\right)}{(2.7\times {10}^{4}K)}$

=0.010

Therefore, ratio of the squares of the speeds before and after the compression is 0.010.