Question

when 5 moles of an ideal gas is expanded at 350k reversibly from 5l to 40l then the entropy change for the process
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Answer 1

• The entropy change for the process is 20.79 cal /K/mol

Given-

• Moles of an ideal gas = 5 moles
• Temperature of an ideal gas = 350 K
• Initial volume (V₁) = 5 lit
• Final volume (V₂) = 40 lit

We know that change in entropy is the ratio of heat and Temperature.

ΔS = Q/T = -W/T , where W is work done. So,

ΔS = -(-2.303 nRT log (V₂/V₁) / T where n is the number of moles, R is the universal gas constant and T is the temperature.

ΔS = 2.303 nR log (V₂/V₁)

By putting the values we get-

ΔS = 2.303 × 5 × 2 × log (40/5)

ΔS = 20.79 cal /K/mol

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