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 isobaric process,
$$\dfrac{V_i}{T_i}=\dfrac{V_f}{T_f}$$
Hence,
$$T_f=\dfrac{V_fT_i}{V_i}$$
where $V_f=4V_i$;
$$T_f=\dfrac{4V_iT_i}{V_i}=4T_i$$
$$\boxed{T_f=4T_i}$$
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a) We know that the rms speed is given by
$$v_{\rm rms}=\dfrac{3kBT}{m}$$
which means that $v_{\rm rms}\propto \sqrt{T}$.
Thus,
$$\boxed{(v_{\rm rms})_f=2v_{\rm rms}}$$
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, $\lambda\propto V$.
$$\boxed{\lambda_f=4\lambda}$$
so, when the volume increased by a factor of 4, the final mean free path will be increased by a factor of 4.
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c) We know that the thermal energy of the gas is given by
$$E_{th}=nC_{\rm V}T$$
which means that $E_{th}\propto T$
Hence, when $T_f=4T$
$$\boxed{(E_{th})_f=4E_{th}}$$
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d) We know that the molar-specific heat at constant volume for a diatomic gas is given by
$$C_{\rm V}=\frac{5}{2}R$$
and $R$ is constant.
Hence, the molar-specific heat at constant volume remains constant.
