Chapter 17, Problem 9STP

Interpretation Introduction

Interpretation:

To calculate the heat change in region B and C for methanol.

Concept Introduction :

When matter transitions from one state (solid, liquid, gas, or plasma), it goes through a phase shift. Both when the pressure on the system changes and when enough energy is added to (or taken away from) the system, these changes take place.

Answer to Problem 9STP

  $\text{37}\text{.9 kJ}$

Explanation of Solution

For methanol, in region B, temperature change from $175.5\text{ K to 337}\text{.2 K}$ .

Initial temperature in $°\text{C}$ = $175.5-273=-\text{97}\text{.5°C}$

Final temperature in $°\text{C}$ = $\text{337}\text{.2}-273=-\text{64}\text{.2°C}$

Molar mass of methanol is $32.04 \text{ g/mol}$ .

No. of mol is 1.

So, mass of methanol is $32.04\text{ g}$

Specific heat for methanol is $2.53\text{J}/\left(\text{g}\cdot °\text{C}\right)$ .

Now calculate heat change in region B

  $\begin{array}{c}\Delta H=mc\Delta t\\ =32.04\text{ g}\times 2.53\text{J}/\left(\text{g}\cdot °\text{C}\right)\times \left(-64.2-\left(-\text{97}\text{.5°C}\right)\right)°\text{C}\\ =32.04\text{ g}\times 2.53\text{J}/\left(\text{g}\cdot °\text{C}\right)\times \text{33}\text{.3}°\text{C}\\ \text{=2,699}\text{.33 J}\\ \text{=2}\text{.7 kJ}\end{array}$

In region C, an only phase transition occurs from liquid to gas. So, heat absorbed will be equal to enthalpy of vaporization.

The enthalpy of vaporization for methanol is $35.2\text{kJ}/\text{mol}$ /

No. of mol given is 1 mol.

Thus, heat absorbed will be $35.2\text{kJ}/\text{mol}$

So, total heat absorbed in region B and C is,

  $\begin{array}{c}\text{Heat absorbed}=\text{2}\text{.7 kJ}+35.2\text{kJ}/\text{mol}\\ =\text{37}\text{.9 kJ}\end{array}$

Thus, total heat absorbed is $\text{37}\text{.9 kJ}$ .