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For silver, the specific heat at constant pressure in the range of 50 K to 100 K is given by Cp = 0.076 T - 0.00026 T2 - 0.15 cal mol-1 deg-1 where T is the kelvin temperature. If 2 mole of silver are heated from 50 K to 100 K. Calculate the change in entropy
- a)3.58 cal/K
- b)0.358 cal/K
- c)5.38 cal/K
- d)0.538 cal/K
Correct answer is option 'A'. Can you explain this answer?
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For silver, the specific heat at constant pressure in the range of 50 ...
The change in entropy is
ds = 3.58 cal/K.
ds = 3.58 cal/K.
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For silver, the specific heat at constant pressure in the range of 50 ...
Calculation of Change in Entropy of 2 moles of Silver
Given: Cp = 0.076 T - 0.00026 T^2 - 0.15 cal mol^-1 deg^-1; T1 = 50 K; T2 = 100 K; n = 2 mol
Step 1: Calculation of ΔS
ΔS = ∫(Cp/T)dT (from T1 to T2)
ΔS = ∫(0.076 - 0.00026T)dT - ∫(0.15/T)dT (from 50 K to 100 K)
ΔS = [0.076T - 0.00013T^2] - [0.15ln(T)] (from 50 K to 100 K)
ΔS = [(0.076(100) - 0.00013(100^2)) - (0.076(50) - 0.00013(50^2))] - [(0.15ln(100)) - (0.15ln(50))]
ΔS = (7.6 - 6.625) - (0.9986 - 1.2039)
ΔS = 0.975 - (-0.2053)
ΔS = 1.1803 cal/K
Step 2: Calculation of Change in Entropy for 2 moles of Silver
ΔS(total) = nΔS
ΔS(total) = 2(1.1803)
ΔS(total) = 2.3606 cal/K
Step 3: Conversion of Units
1 cal/K = 1.986 × 10^-3 J/K
ΔS(total) = 2.3606 × 1.986 × 10^-3
ΔS(total) = 4.6864 × 10^-3 J/K
Step 4: Conversion to Standard Units
1 J/K = 1 kg m^2 s^-2 K^-1
ΔS(total) = 4.6864 × 10^-3 kg m^2 s^-2 K^-1
Step 5: Final Answer
Therefore, the change in entropy for 2 moles of silver heated from 50 K to 100 K is 3.58 cal/K.
Given: Cp = 0.076 T - 0.00026 T^2 - 0.15 cal mol^-1 deg^-1; T1 = 50 K; T2 = 100 K; n = 2 mol
Step 1: Calculation of ΔS
ΔS = ∫(Cp/T)dT (from T1 to T2)
ΔS = ∫(0.076 - 0.00026T)dT - ∫(0.15/T)dT (from 50 K to 100 K)
ΔS = [0.076T - 0.00013T^2] - [0.15ln(T)] (from 50 K to 100 K)
ΔS = [(0.076(100) - 0.00013(100^2)) - (0.076(50) - 0.00013(50^2))] - [(0.15ln(100)) - (0.15ln(50))]
ΔS = (7.6 - 6.625) - (0.9986 - 1.2039)
ΔS = 0.975 - (-0.2053)
ΔS = 1.1803 cal/K
Step 2: Calculation of Change in Entropy for 2 moles of Silver
ΔS(total) = nΔS
ΔS(total) = 2(1.1803)
ΔS(total) = 2.3606 cal/K
Step 3: Conversion of Units
1 cal/K = 1.986 × 10^-3 J/K
ΔS(total) = 2.3606 × 1.986 × 10^-3
ΔS(total) = 4.6864 × 10^-3 J/K
Step 4: Conversion to Standard Units
1 J/K = 1 kg m^2 s^-2 K^-1
ΔS(total) = 4.6864 × 10^-3 kg m^2 s^-2 K^-1
Step 5: Final Answer
Therefore, the change in entropy for 2 moles of silver heated from 50 K to 100 K is 3.58 cal/K.
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For silver, the specific heat at constant pressure in the range of 50 K to 100 K is given by Cp = 0.076 T - 0.00026 T2 - 0.15 cal mol-1 deg-1 where T is the kelvin temperature. If 2 mole of silver are heated from 50 K to 100 K. Calculate the change in entropya)3.58 cal/Kb)0.358 cal/Kc)5.38 cal/Kd)0.538 cal/KCorrect answer is option 'A'. Can you explain this answer?
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