Question

A system of ideal gas at 22°C undergoes an ischoric process with an internal energy decrease of 4.30 × 10 3 4.30×103 J to a final temperature of 16°C. What is the approximate change in entropy?[Hint: use the average temperature]

Answer 1

Answer:

The approximate change in entropy is -14.72 J/K.

Explanation:

Given that,

Temperature = 22°C

Internal energy $U=4.30\times10^{3}\ J$

Final temperature = 16°C

We need to calculate the approximate change in entropy

Using formula of the entropy

$\Delta S=\dfrac{\Delta U}{T}$

Where, $\Delta U$ = internal energy

T = average temperature

Put the value in to the formula

$\Delta S=\dfrac{-4.30\times10^{3}}{\dfrac{22+273+16+273}{2}}$

$\Delta S=-14.72\ J/K$

Hence, The approximate change in entropy is -14.72 J/K.