Hand graded problem 30 moles of O2 gas are confined within a

Hand graded problem. 3.0 moles of O_2 gas are confined within a glass flask with a diaphragm, which allows for the expansion of the gas. The gas is then allowed to expand adiabatically and reversibly. Assume that the O_2 behaves ideally, and that the initial temperature is 298K, and that the vibrational motion contributes only partially to the heat capacity at this temperature R/3 to the heat capacity). Calculate Delta S, Delta S_surr, Delta S_univ. Put a box around your answers. If the process is actually irreversible instead, what can we say about Delta S_univ? Describe how this irreversible process might affect Delta S_univ, even though it is adiabatic.

Solution

case 1:(reversible adiabatic = isentropic process)

in this process entropy change of the system(gas)= 0 (since isentropic process)

change in s = entropy change os system + entropy change of surroundings. = 0 (since no heat interactions with surroundings and moreover process is isentropic)

when ever there is heat generation there wil be entropy change.

case 2 : (irreversible adiabatic)

eventhough the process is adaibatic, i.e.. no heat transfer to surroundings, there will bw increase in the entropy.

this is due to the fact that when the gas expands adiabatically, heat is generated inside the gaswhich will increase the entropy. there will be entropy generation in this case.

note: there will be two modes in which entropy can be increased.

1. entropy transfer through heat

2. entropy generation ( which is always positive or zero )

in case 1 both will be zero. where as in case 2 there will be no entropy transfer but there will be entropy generation.