Answer
a) $3.8\times 10^5\;\rm J$
b) $2.25\;\rm nm$
c) $0\;\rm J$
Work Step by Step
a) The nitrogen is a diatomic gas, so its thermal energy is given by
$$E_{th}=\frac{5}{2}nRT\tag 1$$
Now we need to find the number of moles of nitrogen gas which is given by applying the ideal gas law of
$$PV=nRT$$
$$n=\dfrac{PV}{RT}$$
Plugging into (1);
$$E_{th}=\frac{5}{2} \dfrac{PV}{ \color{red}{\bf\not} R \color{red}{\bf\not} T} \color{red}{\bf\not} R \color{red}{\bf\not} T$$
$$E_{th}=\frac{5}{2}PV $$
Plugging the known;
$$E_{th}=\frac{5}{2}(100\times 1.013\times 10^5)(15000\times 10^{-6}) $$
$$E_{th}=\color{red}{\bf 3.8\times 10^5}\;\rm J$$
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b) The mean free path is given by
$$\lambda=\dfrac{1}{4\sqrt{2}\pi r^2(N/V)}\tag 2$$
and we know from the ideal gas law that $PV=Nk_BT$, so that
$$\dfrac{N}{V}=\dfrac{P}{k_BT}$$
Plugging into (2);
$$\lambda=\dfrac{k_BT}{4\sqrt{2}\pi r^2 P} $$
Plugging the known;
$$\lambda=\dfrac{(1.38\times 10^{-23})(20+273)}{4\sqrt{2}\pi (1\times 10^{-10})^2 (100\times 1.013\times 10^5)} $$
$$\lambda=\bf 2.25\times 10^{-9}\;\rm m=\color{red}{\bf 2.25}\;\rm nm$$
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c) We know that the change in the thermal energy of the gas is given by
$$\Delta E_{th}=nC_{\rm v}\Delta T$$
And the gas expands isothermally which means that the temperature is constant throughout this process. And hence the change in temperature is zero.
Thus,
$$\Delta E_{th}=nC_{\rm v}(0)=\color{red}{\bf 0 }\;\rm J$$
