Answer
a) $1620\;\rm J$
b) $2700\;\rm J$
b) $1549\;\rm m/s$
Work Step by Step
First of all, we need to find the number of moles in the 1.0-g hydrogen gas.
$$n=\dfrac{m}{M_{H_2}}=\dfrac{1}{2(1)}=\bf 0.5\;\rm mol$$
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a) We know that the kinetic energy of one molecule of an ideal gas (such as hydrogen) is given by
$$K=\frac{1}{2}mv_{\rm rms}^2$$
And if we have $N$ molecules, then the total kinetic energy of the gas is then given by
$$K_{tot}=\frac{1}{2}Nmv_{\rm rms}^2$$
We know that the number of molecules is given by $N=nN_{\rm A}$, so
$$K_{tot}=\frac{1}{2}nN_{\rm A}M_{H_2}v_{\rm rms}^2$$
where $M_{H_2}=2M_H$ since $M$ is the atomic mass of the atom.
$$K_{tot}=\frac{1}{2}nN_{\rm A}M_{H}v_{\rm rms}^2$$
$$K_{tot}=\frac{1}{\color{red}{\bf\not}2}nN_{\rm A}(\color{red}{\bf\not}2M_{H})v_{\rm rms}^2$$
$$K_{tot}= nN_{\rm A}M_{H}v_{\rm rms}^2$$
Plugging the known, and do not forget to convert the u to kg.
$$K_{tot}= (0.5)(6.022\times 10^{23}) (1\times 1.661\times 10^{-27})(1800)^2$$
$$K_{tot}= 1620\;\rm J=\color{red}{\bf 1.62\times 10^3}\;\rm J$$
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b)
We know that the thermal energy of a diatomic gas is given by
$$E_{th}=\frac{5}{2}nRT\tag 1$$
And we can find $T$ from the rms speed which is given by
$$v_{\rm rms}=\sqrt{\dfrac{3k_BT}{M_{H_2}}}$$
Hence,
$$T=\dfrac{v_{\rm rms}^2(2M_{H})}{3k_B}\tag 2$$
Plugging into (1);
$$E_{th}=\frac{5}{\color{red}{\bf\not}2}nR \dfrac{v_{\rm rms}^2(\color{red}{\bf\not}2M_{H})}{3k_B}$$
$$E_{th}=5nR \dfrac{v_{\rm rms}^2M_{H}}{3k_B}$$
Plugging the known;
$$E_{th}=5(0.5)(8.31) \dfrac{(1800)^2(1\times 1.661\times 10^{-27})}{3(1.38\times 10^{-23})}$$
$$E_{th}=\color{red}{\bf 2.70\times 10^3}\;\rm J$$
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c) To find the final rms speed, we need to find the final temperature and to find the final temperature we need to find the final thermal energy of the gas.
We know that the change in the thermal energy of the gas is given by
$$\Delta E_{th}=E_{f,th}-E_{i,th}=Q+W$$
Hence,
$$E_{f,th} =Q+W+E_{i,th}$$
Hence,
$$\frac{5}{2}nRT_f =Q+W+E_{i,th}$$
$$T_f=\dfrac{Q+W+E_{i,th}}{\frac{5}{2}nR}$$
Plug this temperature into $v_{\rm rms}=\sqrt{\dfrac{3k_BT_f}{M_{H_2}}}$
$$v_{f,\rm rms}=\sqrt{\dfrac{3k_B }{2M_{H}}\cdot \dfrac{Q+W+E_{i,th}}{\frac{5}{2}nR}}$$
Plugging the known;
$$v_{f,\rm rms}=\sqrt{\dfrac{3(1.38\times 10^{-23}) }{2(1\times 1.661\times 10^{-27})}\cdot \dfrac{(-1200)+(500)+(2700)}{\frac{5}{2}(0.5)(8.31)}}$$
$$v_{f,\rm rms}=\color{red}{\bf 1549}\;\rm m/s$$
