Question

in an isochoric process (i.e., constant volume), heat is added to 10 mol of monoatomic ideal gas whose temperature increases from 273 k to 373 k. what is the entropy change of the gas?

Answer 1

The entropy change for 10 mol of monoatomic ideal gas in an isochoric process with a temperature increase from 273 K to 373 K is 778.5 J/K.

Step-by-step explanation:

The question deals with the concept of entropy change during an isochoric process for a monoatomic ideal gas. Preparing to calculate the entropy change, we use the formula ΔS = nCvln(T2/T1), where ΔS is the entropy change, n is the number of moles of gas, Cv is the molar heat capacity at constant volume, T1 is the initial temperature, and T2 is the final temperature. For a monoatomic ideal gas, Cv is ⅓R, where R is the ideal gas constant. Given that n = 10 mol and the temperature change is from 273 K to 373 K:

Calculate the natural logarithm of the ratio of the final to the initial temperature (ln(373/273)).

Multiply the number of moles (n = 10) by ⅓R and the calculated ln value to get the entropy change (ΔS).

Using R = 8.314 J mol−1 K−1:

ΔS = 10 mol * (⅓ * 8.314 J mol−1 K−1) * ln(373/273)

ΔS ≈ 10 mol * 3 * 8.314 J mol−1 K−1 * 0.312

ΔS ≈ 778.5 J/K

Therefore, the entropy change of the gas is 778.5 J/K.