Answer
See the detailed answer below.
Work Step by Step
First, we need to find the final temperature of the gas in terms of its initial temperature so we can easily answer the following questions.
For an adiabatic process,
$$T_iV_i^{\gamma-1}= T_fV_f^{\gamma-1}$$
Hence,
$$T_f=T_i\left[\dfrac{V_i}{V_f}\right]^{\gamma-1}$$
where $V_f=\frac{1}{8}V_i$, and $\gamma=\frac{5}{3}$ for monatomic gas.
Hence,
$$T_f=T_i\left[\dfrac{8V_i}{V_i}\right]^{\frac{5}{3}-1}=(8)^{\frac{2}{3}}$$
$$\boxed{T_f=4T_i}$$
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a) We know that the rms speed is given by
$$v_{\rm ms}=\sqrt{\dfrac{3k_BT}{m}}$$
Hence, $v_{\rm ms}\propto \sqrt{T}$, so when $T=4T_i$,
$$\boxed{(v_{\rm rms})_f=2v_{\rm rms}}$$
The rms speed is increased by a factor of 2.
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b) We know that the mean free path is given by
$$\lambda=\dfrac{1}{4\sqrt{2}\pi r^2 (N/V)}=\dfrac{V}{4\sqrt{2}\pi N r^2}$$
Hence, $\lambda\propto V$
$$\boxed{\lambda_f=\frac{1}{8}\lambda} $$
Thus the final mean free path will be decreased by a factor of $\frac{1}{8}$.
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c) We know that the thermal energy of the gas is given by
$$E_{th}=nC_{\rm V}T$$
Hence, $E_{th}\propto T$
and hence,
$$(E_{th})_f=nC_{\rm V}(4T)$$
$$\boxed{(E_{th})_f=4 E_{th}}$$
The final thermal energy of the gas will be increased by a factor of 4.
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d) We know that the molar-specific heat at constant volume for a monatomic gas is given by
$$C_{\rm V}=\frac{3}{2}R$$
and $R$ is constant.
Hence, the molar-specific heat at constant volume remains constant.