#### Answer

For every two molecules of $NH_3$ consumed, 4 molecules of gas are produced. Therefore, the resulting gas will have twice the volume as there is now twice the number of molecules as previously.
The total pressure increases to twice the previous pressure.
The partial pressure of $N_2$ is half the pressure of $NH_3$ and one-third the pressure of $H_2$. The partial pressure of $H_2$ is 50% more than the partial pressure of $NH_3$ and three times the partial pressure of $N_2$.

#### Work Step by Step

The chemical equation for the decomposition of $NH_3$ is as follows:
$$2NH_{3(g)}\Rightarrow N_{2(g)} + 3H_{2(g)}$$
In this reaction, two moles of gas are converted into four moles of gas. In other words, we are doubling the number of moles of gas.
As a result, if temperature and pressure are held constant, the volume doubles by Avogadro's law.
Furthermore, by the ideal gas law, because volume and temperature are constant, the total pressure is twice the previous pressure as we have twice the number of moles of gas.
With the decomposition of 2 moles of $NH_3$, one mole of $H_2$ is produced and 3 moles of $H_2$ is produced. Therefore, the partial pressure of $H_2$ is one-half the partial pressure of $NH_3$ and one-third the partial pressure of $H_2$. Similarly, the partial pressure of $H_2$ is 50% greater than the partial pressure of $NH_3$ and three times the partial pressure of $N_2$.