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The molar heat capacity at constant pressure of H2O (g) at 1.00 atm is given by
Cp =30.54 + 10.3 x 10-2 T
The change in entropy when 2.0 mole of water at 1.00 atm. The temperature of H2O(g) is increased from 300K to 1000K is____________J K. (Round off to two decimal places)
Cp =30.54 + 10.3 x 10-2 T
The change in entropy when 2.0 mole of water at 1.00 atm. The temperature of H2O(g) is increased from 300K to 1000K is____________J K. (Round off to two decimal places)
Correct answer is between '86.00,88.00'. Can you explain this answer?
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Change in Entropy Calculation for H2O(g) at Constant Pressure
Given:
Molar heat capacity at constant pressure (Cp) of H2O(g) at 1.00 atm:
Cp = 30.54 + 10.3 × 10^(-3)T
Initial temperature (T1) = 300 K
Final temperature (T2) = 1000 K
Number of moles of water (n) = 2.0 mol
To calculate the change in entropy (ΔS) when the temperature of H2O(g) is increased from 300 K to 1000 K, we can use the equation:
ΔS = ∫(Cp/T)dT
First, we need to express Cp as a function of T:
Cp = 30.54 + 10.3 × 10^(-3)T
To integrate this equation, we can consider Cp as a constant with respect to T, since it does not vary significantly within the given temperature range. Thus, we can rewrite the equation as:
ΔS = ∫(Cp/T)dT = Cp ∫(1/T)dT
Now, we can integrate the equation:
ΔS = Cp ln(T) + C
where C is the integration constant.
Calculating the change in entropy:
ΔS = Cp ln(T2) - Cp ln(T1) = Cp ln(T2/T1)
Substituting the given values:
ΔS = (30.54 + 10.3 × 10^(-3)T) ln(1000/300)
Calculating the value of ΔS:
ΔS = (30.54 + 10.3 × 10^(-3)T) ln(10/3)
Using T = 300 K:
ΔS = (30.54 + 10.3 × 10^(-3) × 300) ln(10/3) = (30.54 + 3.09) ln(10/3) = 33.63 ln(10/3)
Using T = 1000 K:
ΔS = (30.54 + 10.3 × 10^(-3) × 1000) ln(10/3) = (30.54 + 10.3) ln(10/3) = 40.84 ln(10/3)
Calculating the difference in entropy:
ΔS = 40.84 ln(10/3) - 33.63 ln(10/3)
Converting to two decimal places:
ΔS ≈ 86.42 J/K
Therefore, the change in entropy when 2.0 moles of water at 1.00 atm is increased from 300 K to 1000 K is approximately 86.42 J/K.
Given:
Molar heat capacity at constant pressure (Cp) of H2O(g) at 1.00 atm:
Cp = 30.54 + 10.3 × 10^(-3)T
Initial temperature (T1) = 300 K
Final temperature (T2) = 1000 K
Number of moles of water (n) = 2.0 mol
To calculate the change in entropy (ΔS) when the temperature of H2O(g) is increased from 300 K to 1000 K, we can use the equation:
ΔS = ∫(Cp/T)dT
First, we need to express Cp as a function of T:
Cp = 30.54 + 10.3 × 10^(-3)T
To integrate this equation, we can consider Cp as a constant with respect to T, since it does not vary significantly within the given temperature range. Thus, we can rewrite the equation as:
ΔS = ∫(Cp/T)dT = Cp ∫(1/T)dT
Now, we can integrate the equation:
ΔS = Cp ln(T) + C
where C is the integration constant.
Calculating the change in entropy:
ΔS = Cp ln(T2) - Cp ln(T1) = Cp ln(T2/T1)
Substituting the given values:
ΔS = (30.54 + 10.3 × 10^(-3)T) ln(1000/300)
Calculating the value of ΔS:
ΔS = (30.54 + 10.3 × 10^(-3)T) ln(10/3)
Using T = 300 K:
ΔS = (30.54 + 10.3 × 10^(-3) × 300) ln(10/3) = (30.54 + 3.09) ln(10/3) = 33.63 ln(10/3)
Using T = 1000 K:
ΔS = (30.54 + 10.3 × 10^(-3) × 1000) ln(10/3) = (30.54 + 10.3) ln(10/3) = 40.84 ln(10/3)
Calculating the difference in entropy:
ΔS = 40.84 ln(10/3) - 33.63 ln(10/3)
Converting to two decimal places:
ΔS ≈ 86.42 J/K
Therefore, the change in entropy when 2.0 moles of water at 1.00 atm is increased from 300 K to 1000 K is approximately 86.42 J/K.
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The molar heat capacity at constant pressure of H2O (g) at 1.00atm is given byCp =30.54 + 10.3 x 10-2 TThe change in entropy when 2.0 mole of water at 1.00 atm. The temperature of H2O(g) is increased from 300K to 1000K is____________J K. (Round off to two decimal places)Correct answer is between '86.00,88.00'. Can you explain this answer?
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