# Changing conditions of gaseous mixture

A gaseous mixture of $$\ce{NO2}$$ and $$\ce{N2O4}$$ is sealed in 4 identical vessels with movable pistons. When the conditions are modified as described in 1) to 4), in which condition does the molar ratio of $$\ce{NO2}$$ to $$\ce{N2O4}$$ remain unchanged after a long time enough to reach equilibrium?
1) Total pressure is increased by moving the piston at a constant temperature.
2) Temperature is increased while keeping the total pressure constant by moving the piston.
3) Additional $$\ce{NO2}$$ is introduced from outside while keeping the total pressure constant by moving the piston at a constant temperature.
4) Additional $$\ce{N2}$$ is introduced from outside while keeping the total pressure constant by moving the piston at a constant temperature.

My attempt: I think when we add $$\ce{NO2}$$, the equilibrium will shift to the right so the concentration of those two will changed. I answered (4) because $$\ce{N2}$$ is unreactive, making the concentration of both compounds the same. Please help why the answer 3 is correct and others are wrong.
• Because when we balance the equation we have $\ce{2NO2}$ => $\ce{N2O4}$. Although the pressure remains constant, since choice (3) introduced $\ce{NO2}$ (increase in concentration), the system will counteract the change by shifting equilibrium to the right. – Trey Anupong Jun 21 '19 at 16:42
• Thank you for your hint. I have looked closer again and yes, both before and after addition will give the same equilibrium constant, meaning the molar ratio of those two compounds is equal so (3) is correct. I am also interested in the choices that are wrong. The equilibrium constant will change only when the temperature changes if I understand it correctly so number (2) is indeed incorrect. However, I am still struggling why number (1) and (4) are incorrect at a constant temperature. Like in number (4), adding inert $\ce{N2}$ would not affect partial pressure of those two compounds. – Trey Anupong Jun 21 '19 at 19:16