Question:
The specific heat \(C_p(T)\) of one mole of a material as a function of temperature \(T\) is given as \(C_p(T) = AT + BT^3\), where \(A = 0.695\) mJ.mol\(^{-1}\).K\(^{-2}\) and \(B = 0.045\) mJ.mol\(^{-1}\).K\(^{-4}\). When \(T\) is changed from 1 K to 10 K at constant pressure, the change in entropy \(\Delta S\) in mJ.mol\(^{-1}\).K\(^{-1}\) (rounded off to one decimal place) is
The specific heat \(C_p(T)\) of one mole of a material as a function of temperature \(T\) is given as \(C_p(T) = AT + BT^3\), where \(A = 0.695\) mJ.mol\(^{-1}\).K\(^{-2}\) and \(B = 0.045\) mJ.mol\(^{-1}\).K\(^{-4}\). When \(T\) is changed from 1 K to 10 K at constant pressure, the change in entropy \(\Delta S\) in mJ.mol\(^{-1}\).K\(^{-1}\) (rounded off to one decimal place) is
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Hint:
Divide \(C_p\) by \(T\) first, then integrate the resulting polynomial from 1 K to 10 K.
Divide \(C_p\) by \(T\) first, then integrate the resulting polynomial from 1 K to 10 K.
Updated On: Jul 28, 2026
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Correct Answer: 21.2
Solution and Explanation
Step 1: Understanding the Concept:
Entropy change at constant pressure comes from integrating \(C_p/T\) over the temperature range, since \(dS = \dfrac{C_p}{T}dT\) at constant pressure.
Step 2: Key Formula or Approach:
Divide the given \(C_p(T) = AT + BT^3\) by \(T\):
\[ \frac{C_p}{T} = A + BT^2 \]
Then integrate from \(T_1 = 1\) K to \(T_2 = 10\) K:
\[ \Delta S = \int_{1}^{10} (A + BT^2)\, dT = A(T_2 - T_1) + \frac{B}{3}(T_2^3 - T_1^3) \]
Step 3: Detailed Explanation:
With \(A = 0.695\) and \(B = 0.045\) (both in the mJ.mol\(^{-1}\).K units given), and \(T_2 - T_1 = 9\), \(T_2^3 - T_1^3 = 1000 - 1 = 999\):
\[ \Delta S = 0.695 \times 9 + \frac{0.045}{3} \times 999 \]
\[ \Delta S = 6.255 + 0.015 \times 999 = 6.255 + 14.985 \]
\[ \Delta S = 21.24 \text{ mJ.mol}^{-1}.\text{K}^{-1} \]
Final Answer:
Rounded to one decimal place, the entropy goes up by 21.2 mJ per mole per kelvin as the sample warms from 1 K to 10 K. \[ \boxed{\Delta S = 21.2 \text{ mJ.mol}^{-1}.\text{K}^{-1}} \]
Entropy change at constant pressure comes from integrating \(C_p/T\) over the temperature range, since \(dS = \dfrac{C_p}{T}dT\) at constant pressure.
Step 2: Key Formula or Approach:
Divide the given \(C_p(T) = AT + BT^3\) by \(T\):
\[ \frac{C_p}{T} = A + BT^2 \]
Then integrate from \(T_1 = 1\) K to \(T_2 = 10\) K:
\[ \Delta S = \int_{1}^{10} (A + BT^2)\, dT = A(T_2 - T_1) + \frac{B}{3}(T_2^3 - T_1^3) \]
Step 3: Detailed Explanation:
With \(A = 0.695\) and \(B = 0.045\) (both in the mJ.mol\(^{-1}\).K units given), and \(T_2 - T_1 = 9\), \(T_2^3 - T_1^3 = 1000 - 1 = 999\):
\[ \Delta S = 0.695 \times 9 + \frac{0.045}{3} \times 999 \]
\[ \Delta S = 6.255 + 0.015 \times 999 = 6.255 + 14.985 \]
\[ \Delta S = 21.24 \text{ mJ.mol}^{-1}.\text{K}^{-1} \]
Final Answer:
Rounded to one decimal place, the entropy goes up by 21.2 mJ per mole per kelvin as the sample warms from 1 K to 10 K. \[ \boxed{\Delta S = 21.2 \text{ mJ.mol}^{-1}.\text{K}^{-1}} \]
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