Question 18

[naveen.kri8901]

Can you explain in more detail how you figure out that the atmospheric pressure is greater than the reaction pressure like it is stated in the answer explanation?

Thanks.

[hkidd3099]

Well, notice that the question states that when the reaction was started, the water levels were equal on both sides, which indicates that there was no difference between atmospheric pressure and the pressure within the reaction vessel - the reaction was at atmospheric pressure level. However, as the reaction occurs, the pressure inside the reaction vessel must have decreased, because the reaction was forming fewer moles of gas. Now, we can assume that the pressure in the reaction vessel is less than atmospheric pressure, and the water level will thus shift accordingly up on the left tube, since atmospheric pressure serves to "push" the water down on the right side of the tube.

[christophe1685]

The question also states that the temperature increases during the reaction. Based on PV=nRT there isn't a change in V or R, however since n decreases and T increases we really can't determine what happens to P without knowing what the temperature change is. Since there isn't a 'don't know' option you're stuck with assuming T is constant (which the question states isn't the case).

[nedaa.asba6809]

If you want to calculate the pressure in the container, you have to take into account the atmospheric pressure and that's done by comparing the height (∆h) in the two arms of the manometer, where
∆P = pg∆h.

[ames.rober3649]

what about the fact that the vapor pressure of ether is less than that of pure water on the other side?

[btdumford5156]

Closed spontaneous reaction that increase the temperature of the system, and no increase in pressure?

[dltkdgn896512]

snaps... i thought of this q as a discrete, silly me

what you have to do is to consider that 3 mole of gas changed to 2 moles of gas... thus less pressure due to less amount of moles of gas.