Question

A force of 22.7 N stretches an elastic band at room temperature. The rate at which its entropy changes as it stretches is about _____ J/Km. Round your answer to 3 decimal places.

Answer 1

The value is
`(\Delta S )/( L) = - 0.0721 \ J / km`

Step-by-step explanation:

From the question we are told that

The force is
`F = 22.7 \ N`

The value of room temperature is
`T = 298 \ K`

Generally the rate at which its entropy changes as it stretches is mathematically represented as

`(\Delta S )/( L) = - (F)/(T)`

=>
`(\Delta S )/( L) = - (21.5)/( 298 )`

=>
`(\Delta S )/( L) = - 0.0721 \ J / km`