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  • Given <img alt="\mathrm{C_p=20+0.02 T\, \, (in \left.\mathrm{J} / \mathrm{mol} \cdot \mathrm{K}\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BC_p%3D20&plus;0.02%20T%

Given \mathrm{C_p=20+0.02 T\, \, (in \left.\mathrm{J} / \mathrm{mol} \cdot \mathrm{K}\right)}, where T is the temperature in Kelvin. Calculate the change in enthalpy

\mathrm{(\Delta H)} for a process in which the temperature of 2 moles of the substance increases from \mathrm{300 \mathrm{~K}\, \, to \, \, 500 \mathrm{~K}.}

Option: 1

12 KJ


Option: 2

15 KJ


Option: 3

19 KJ


Option: 4

10 KJ


Answers (1)

Step 1: Calculate \mathrm{C_p} at both temperatures

                                   \mathrm{ \begin{aligned} & C_{p 1}=20+0.02 \cdot 300=26 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \\\\ & C_{p 2}=20+0.02 \cdot 500=30 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \end{aligned} }
Step 2: Calculate the change in temperature \mathrm{\Delta T=T_2-T_1=500 \mathrm{~K}- 300 \mathrm{~K}=200 \mathrm{~K}}

Step 3: Calculate the change in enthalpy using the formula \mathrm{\Delta H=n \cdot C_p \cdot \Delta T}

                             \mathrm{ \Delta H=2 \mathrm{~mol} \cdot(30 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}) \cdot 200 \mathrm{~K}=12000 \mathrm{~J}=12 \mathrm{~kJ} }
Therefore, the change in enthalpy \mathrm{(\Delta H)} for the given process is \mathrm{12 \mathrm{~kJ}}. Therefore, the correct option is \mathrm{\mathbf{A}.}

Posted by

Ramraj Saini

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