In the Are You Wondering \(7-1\) box, the temperature variation of enthalpy is discussed, and the equation \(q_{P}=\) heat capacity \(\times\) temperature change \(=C_{P} \times \Delta T\) was introduced to show how enthalpy changes with temperature for a constant-pressure process. Strictly speaking, the heat capacity of a substance at constant pressure is the slope of the line representing the variation of enthalpy (H) with temperature, that is $$C_{P}=\frac{d H}{d T} \quad(\text { at constant pressure })$$ where \(C_{P}\) is the heat capacity of the substance in question. Heat capacity is an extensive quantity and heat capacities are usually quoted as molar heat capacities \(C_{P, \mathrm{m}},\) the heat capacity of one mole of substance; an intensive property. The heat capacity at constant pressure is used to estimate the change in enthalpy due to a change in temperature. For infinitesimal changes in temperature, $$d H=C_{p} d T \quad(\text { at constant pressure })$$ To evaluate the change in enthalpy for a particular temperature change, from \(T_{1}\) to \(T_{2}\), we write $$\int_{H\left(T_{1}\right)}^{H\left(T_{2}\right)} d H=H\left(T_{2}\right)-H\left(T_{1}\right)=\int_{T_{1}}^{T_{2}} C_{P} d T$$ If we assume that \(C_{P}\) is independent of temperature, then we recover equation (7.5) $$\Delta H=C_{P} \times \Delta T$$ On the other hand, we often find that the heat capacity is a function of temperature; a convenient empirical expression is $$C_{P, \mathrm{m}}=a+b T+\frac{c}{T^{2}}$$ What is the change in molar enthalpy of \(\mathrm{N}_{2}\) when it is heated from \(25.0^{\circ} \mathrm{C}\) to \(100.0^{\circ} \mathrm{C} ?\) The molar heat capacity of nitrogen is given by$$C_{P, \mathrm{m}}=28.58+3.77 \times 10^{-3} T-\frac{0.5 \times 10^{5}}{T^{2}} \mathrm{JK}^{-1} \mathrm{mol}^{-1}$$

Step by step solution

01

Convert temperatures to Kelvin

To start with, convert the provided temperatures from Celsius to Kelvin. This is done by adding 273.15 to the Celsius temperature. Let \( T_1 = 25.0^{\circ}C = 298.15K \) and \( T_2 = 100.0^{\circ}C = 373.15K \).

02

Define the molar heat capacity function

The given formula for molar heat capacity of nitrogen is \( C_{P,m} = 28.58 + 3.77 \times 10^{-3} T - \frac{0.5 \times 10^{5}}{T^2} JK^{-1}mol^{-1} \), where \( T \) is the temperature in Kelvin. This will be used later for finding the integral.

03

Use Calculus to find ΔH

The ΔH from temperature \( T_1 \) to \( T_2 \) can be found by evaluating the integral \( \int_{T1}^{T2} C_{P, m} dT \). This integral will give us the molar change in enthalpy.

04

Evaluate the integral

Now all that is left is to evaluate this integral. When this is done, it will give us the change in molar enthalpy for the given temperatures. Please note that evaluation of the integral requires knowledge of calculus and specific integral formulas.

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