Answer
See the detailed answer below.
Work Step by Step
a) The two gases, $\rm O_2$, and $\rm N_2$ are diatomic gases.
Then the thermal energy of the gas inside the room is given by
$$E_{th}=\frac{5}{2}(n_{\rm O_2}+n_{\rm H_2})RT\tag 1$$
Now we need to find the number of moles of each gas, so we can use the ideal gas law.
$$PV=nRT$$
Hence,
$$n=\dfrac{PV}{RT}$$
The two gases are filling the same volume under the same STP conditions and both of them are diatomic gas.
Thus,
$$n_{tot}=n_{\rm O_2}+n_{\rm N_2}=\dfrac{(1.013\times 10^5)(2\times 2\times 2)}{(8.31)(0+273)}$$
$$n_{\rm O_2}=n_{\rm N_2}=\bf 357\;\rm mol$$
Plugging into (1);
$$E_{th}=\frac{5}{2}(357)RT=\frac{5}{2}(357)(8.31)(0+273)$$
$$E_{th}=\color{red}{\bf 2.025\times 10^6}\;\rm J$$
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b) The needed energy to raise the ball 1-m is given by
$$\Delta E_{ball}=mgy=(1)(9.8)(1)=9.8\;\rm J$$
The ball needs about 9.8 J of kinetic energy to reach the height of 1-m and then stops.
Hence, the fraction of thermal energy is given by
$$\dfrac{\Delta E_{ball}}{E_{th}}=\dfrac{9.8}{2.025\times 10^6}$$
$$\dfrac{\Delta E_{ball}}{E_{th}}=\color{red}{\bf 4.84\times 10^{-6}}$$
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c) The final thermal energy of the gas after launching the ball is given by
$$E_{th,f}=E_{th}-\Delta E$$
And hence,
$$\frac{5}{2}nRT_f=E_{th}-\Delta E$$
Thus,
$$T_f=\dfrac{\frac{5}{2}nRT-\Delta E_{ball}}{\frac{5}{2}nR}$$
Plugging the known;
$$T_f=\dfrac{\frac{5}{2}(357)(8.31)(273)-(9.8)}{\frac{5}{2}(357)(8.31)}=\bf 272.999\;\rm K$$
Hence,
$$\Delta T=T_f-T_i=272.999-273=\color{red}{\bf -0.001}\;\rm K$$
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d) It will never happen because both, the ball and the room air, are in thermal equilibrium. Hence, there is no heat exchange.
And even if the ball was initially at less temperature, the heat transferred from the air to it will be just for neutralizing their temperatures.
