Question

In an isochoric process, heat is added to 10 mol of monoatomic ideal gas whose temperature increases from 273 to 373 K. What is the entropy change of the gas?

a) 29.1 J/K, -
b) -29.1 J/K, -
c) 29.1 J/K, 29.1 J/K
d) -29.1 J/K, 29.1 J/K

Answer 1

The entropy change of 10 mol of a monoatomic ideal gas in an isochoric process where the temperature increases from 273 to 373 K is 29.1 J/K. The correct answer is option A

Step-by-step explanation:

The question pertains to calculating the entropy change of a monoatomic ideal gas undergoing an isochoric process. In such a process, the volume remains constant, and therefore no external work is done by the gas. The entropy change ΔS is given by the formula ΔS = nCvln(T2/T1), where n is the number of moles, Cv is the molar heat capacity at constant volume for a monoatomic gas (⅓R, where R is the ideal gas constant), and T1 and T2 are the initial and final temperatures respectively.

In this case, with n = 10 mol, T1 = 273 K, and T2 = 373 K, and given that Cv for a monoatomic ideal gas is ⅓R (≈ 12.47 J/(mol·K)), the entropy change ΔS is:

ΔS = 10 mol · 12.47 J/(mol·K) · ln(373/273)

ΔS = 29.1 J/K