Question

A monatomic gas is adiabatically compressed to $\frac{1}{8}$of its initial volume. Does each of the following quantities change? If so, does it increase or decrease, and by what factor? If not, why not?

a. The $rms$speed.

b. The mean free path.

c. The thermal energy of the gas.

d. The molar specific heat at constant volume.

Short answer

a. The $rms$speed increases by a factor of $2$.

b. The mean free path decreases by a factor of $8$.

c. The thermal energy increases by a factor of $4$.

d. The molar specific heat at constant volume of the gas remains constant

Step-by-step solution

Introduction (part a)

a. The following temperature and volume relationship holds for the adiabatic process:

$T{V}^{\gamma -1}=const$

Therefore

$T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$

This yields

$\frac{T_2}{T_1}={\left(\frac{V_1}{V_2}\right)}^{\gamma -1}$

in the equation , it is given that $\raisebox{1ex}{$V_1$}\!\left/ \!\raisebox{-1ex}{$V_2$}\right.=8$so

$\frac{T_2}{T_1}=8^{\gamma -1}$

The adiabatic exponent for monoatomic gases is $\gamma =5/3$so

$\frac{T_2}{T_1}=8^{\frac{5}{3}-1}=8^{\frac{2}{3}}=(\sqrt[3]{8}{)}^2=4$

As a result, the temperature rises by a factor of ten in this procedure.$4$. Remember that $v_{rms}$is proportional to $\sqrt{T}$so we have

$\frac{v_{\mathrm{rms}2}}{v_{\mathrm{rms}1}}=\sqrt{\frac{T_2}{T_1}}=\sqrt{4}=2$

i.e. The $rms$speed increases by a factor of $2$.

Step 2:Explanation (part b)

b. The mean free path is given by eq. $(20.3)$in the book

$\lambda =\frac{1}{4\sqrt{2}\pi \frac{N}{V}{r}^2}=\frac{V}{4\sqrt{2}\pi N{r}^2}$

where $r$is the effective radius of a molecule, $V$is the volume of the gas and $N$is the number of molecules. All of the quantities in the formula for$\lambda$ are constant in this process except the volume. Since$\lambda \infty V$decreases by a factor of $8$, also $\lambda$ decreases by a factor of $8$.

Step 3:Thermal energy of a monoatomic gas (part c)

c. A monoatomic gas's thermal energy is given by

$E_{\mathrm{th}}=n{C}_{V}T$

Both the number of moles $n$and the molar specific heat at constant volume $C_v$are constant in this process so the only thing that changes is the temperature. Since $E_{th}\infty T$, since we have found that the temperature increases by a factor of $4$then $E_{th}$also increases by a factor of $4$.

Step 4:Heat of a monoatomic gas (part d)

d. The molar specific heat of a monoatomic gas is just a constant equal to

$C_V=\frac{3}{2}R$

so it doesn't change in this process.