Question

Nitrogen gas is to be heated from 373 K to 523 K. The equation for Cp is : A+BT+CT2+DT3 The specific heat constants for Nitrogen are: A=31.15,B=−1.357×10−2,C=26.796×10−6 and D=−1.168×10−8 These constants give your specific heat in Jmol−1 K−1. Determine the mean heat capacity for this temperature range. Give your answer in J mol−1K−1 to 2 decimal places.

Answer 1

Step-by-step explanation: To determine the mean heat capacity for the given temperature range, we can use the equation for Cp: Cp = A + BT + CT^2 + DT^3, where Cp is the specific heat capacity in Jmol^−1 K^−1, T is the temperature in K, and A, B, C, and D are the specific heat constants for nitrogen.

Given:

A = 31.15 Jmol^−1 K^−1

B = −1.357 × 10^−2 Jmol^−1 K^−2

C = 26.796 × 10^−6 Jmol^−1 K^−3

D = −1.168 × 10^−8 Jmol^−1 K^−4

Initial temperature (T1) = 373 K

Final temperature (T2) = 523 K

To find the mean heat capacity, we need to integrate the Cp equation with respect to temperature from T1 to T2, and then divide the result by the temperature range (T2 - T1).

Let's calculate the integral step by step:

∫(A + BT + CT^2 + DT^3) dT

Integrating each term separately:

∫A dT = A∫dT = AT

∫BT dT = B∫TdT = B(T^2/2)

∫CT^2 dT = C∫T^2 dT = C(T^3/3)

∫DT^3 dT = D∫T^3 dT = D(T^4/4)

Now, we'll substitute the limits of integration:

∫(A + BT + CT^2 + DT^3) dT = A∫dT + B∫TdT + C∫T^2 dT + D∫T^3 dT

= AT + B(T^2/2) + C(T^3/3) + D(T^4/4)

Evaluating the integral from T1 to T2:

∫(A + BT + CT^2 + DT^3) dT = [AT + B(T^2/2) + C(T^3/3) + D(T^4/4)](T2 - T1)

Finally, to find the mean heat capacity, we divide the result by the temperature range (T2 - T1):

Mean heat capacity = [AT + B(T^2/2) + C(T^3/3) + D(T^4/4)](T2 - T1) / (T2 - T1)

Simplifying the expression:

Mean heat capacity = AT + B(T^2/2) + C(T^3/3) + D(T^4/4)

Now, substitute the values of A, B, C, D, T1, and T2 into the equation and calculate the mean heat capacity. Remember to round your final answer to 2 decimal places.