Question

The entropy change when two moles of ideal monoatomic gas is heated from 200 to 300 ∘ C reversibly and isochorically -

Answer 1

The entropy change when two moles of

ideal monoatomic gas is heated from 200 to

300⁰C reversibly and isochorically

answer: hope it helps

Answer 2

Answer: The entropy change is 4.78 J/K

Explanation:

The chemical equation used to calculate entropy change for reversible isochoric process follows:

$\Delta S=nC_v\ln \frac{T_2}{T_1}$

where,

$\Delta S$ = Entropy change = ?

n = number of moles of gas = 2 moles

$C_v$ = specific heat capacity at constant volume = $\frac{3}{2}R$ (for monoatomic gas)

R = Gas constant = 8.314 J/mol.K

$T_1$ = initial temperature = $200^oC=[200+273]K=473K$

$T_2$ = final temperature = $300^oC=[300+273]K=573K$

Putting values in above equation, we get:

$\Delta S=2mol\times \frac{3}{2}\times 8.314J/mol.K\times \ln \frac{573}{473}\\\\\Delta S=4.78J/K$

Hence, the entropy change is 4.78 J/K