Question

Two flasks of equal volume are connected by a narrow tube are at 27°C and contains 0.7 mole of hydrogen gas at 0.5 atm pressure. One of the flasks is then immersed in to a bath kept at 127°C while the other remain at 27°C . Calculate the final pressure and number of moles of hydrogen gas in each flask

Answer 1

Answer:

Final pressure in each flask is 0.5714 atm.

The number moles of hydrogen in one of the flask is 0.3 mol and in other flask is 0.4 mol.  

Explanation:

Given Data:

Two flasks of equal volume are at initial temperature 27°C, initial pressure 0.5 atm and contains 0.7 moles of H₂.

To find: final pressure in each flask

Let the two flasks be “a” & “b”. Also, let flask “a” be immersed in a bath at temperature, Ta = 127°C = 127 + 273.15 = 400.15 K and flask “b” remains the same as initial temperature i.e., temperature, Tb = 27°C + 273.15 = 300.15 K . No. of mole for flask a and b be “na” & “nb” respectively.

Also, the final pressure in both the flask Pa and Pb will be same i.e., Pa = Pb = P and also Va = Vb = V.  

Using Ideal Gas Law for the combined system initially, we have

PV = nRT

⇒ 0.5 * 2V = 0.7 * R * 300.15 …[both the flasks have the same volume]

⇒ V = 210.105 R ….. (i)

For flask a :

Pa * Va = na * R * Ta

⇒ P * V = na * R * Ta

⇒ P * 210.105 R = na * R * 400.15 ….. [ value of V from (i)]

⇒ na = 0.525P ….. (ii)

For flask b:

Pb * Vb = nb * R * Tb

⇒ nb = (Pb Vb) / (R Tb)

⇒ nb = (P * 210.105 R) / (R * 300.15 )

⇒ nb = 0.7P ….. (iii)

We know,  

Total no.of moles initially was = na + nb

Substituting values of na and nb from (ii) & (iii)

⇒ 0.7 = 0.525P + 0.7P

⇒ 0.7 = 1.225 P

⇒ P = 0.7 / 1.225 = 0.5714 atm

∴ na = 0.525P = 0.525 * 0.5714 = 0.299 mol ≈ 0.3 mol

And,

nb = 0.7P = 0.7 * 0.5714 = 0.399 mol ≈ 0.4 mol