Question

The same reaction takes place in two different vessels, which are initially at atmospheric pressure (1 atm). The first vessel is rigid and does not expand with change in pressure; the second vessel will expand to maintain a pressure of 1 atm inside. Over the course of a reaction, the second vessel expands by 1 liter.

A) What is the amount of PV work done by the reaction for each of the vessels?

B) Which vessel’s content has a higher enthalpy?

C) Based on the answer to part b, which vessel would you expect to have a higher temperature?

Answer 1

Solution:

Part (A)

In general the work done by the vessel is calculated by using the formulae,

w = 2.303 nRT log (V₂/V₁) {consider initially at standard condition of T = 25^o C

for vessel 1,

w = 2.303 (1) (8.314) log (1/1) = 0

the work done by vessel 1 is void

for vessel 2 change in volume from 1 lt to 2 lt,

w = 2.303 (1) (8.314)log(2/1) = 5.763

   0.3010 done while comparing with vessel 1

Part (B)

By general we know work is related with enthalpy, on considering enthalpy the energy radiated to atmosphere is high in vessel 2 ao vessel 2 containis higher enthalpy

w = q = enthalpy

Part (C)

Vessel 1 will have high temperature as it not expanded instead releases energy in terms of temperature to the atmosphere, while vessel 2 expands by spending all the observed energy in terms of expansion so temperature rate is less in vessel 2 on comparison with vessel 1.