Certainly! Let's solve the problem step-by-step:
### Given Values
- Enthalpy change, [tex]\( \Delta H = 40.79 \text{ kJ/mol} \)[/tex] (Note: 1 kJ = 1000 J, so [tex]\( \Delta H = 40790 \text{ J/mol} \)[/tex])
- Gas constant, [tex]\( R = 8.314 \text{ J/(mol·K)} \)[/tex]
- Initial temperature, [tex]\( T_1 = 373 \text{ K} \)[/tex]
- Initial pressure, [tex]\( P_1 = 79.8 \text{ kPa} \)[/tex]
- Final pressure, [tex]\( P_2 = 101.3 \text{ kPa} \)[/tex].
### Step-by-Step Calculation
1. Convert [tex]\(\Delta H\)[/tex] to J/mol (if not already in this unit):
[tex]\[
\Delta H = 40.79 \text{ kJ/mol} = 40790 \text{ J/mol}
\][/tex]
2. Calculate the term inside the logarithm:
[tex]\[
\frac{P_1}{P_2} = \frac{79.8 \text{ kPa}}{101.3 \text{ kPa}}
\][/tex]
3. Evaluate the logarithm:
[tex]\[
\ln \left(\frac{79.8}{101.3}\right)
\][/tex]
4. Calculate [tex]\( \frac{1}{T_1} \)[/tex]:
[tex]\[
\frac{1}{373 \text{ K}}
\][/tex]
5. Combine the terms inside the parentheses:
[tex]\[
\frac{1}{373 \text{ K}} - \ln \left(\frac{79.8}{101.3}\right)
\][/tex]
6. Evaluate the term:
[tex]\[
\left(\frac{1}{373 \text{ K}} - \ln \left(\frac{79.8}{101.3}\right)\right)
\][/tex]
Based on the given solution, this term evaluates to:
[tex]\[
\left(\frac{1}{373 \text{ K}} - \ln \left(\frac{79.8}{101.3}\right)\right) = 0.2412438719463277
\][/tex]
7. Calculate the inverse of the term:
[tex]\[
\left(0.2412438719463277\right)^{-1}
\][/tex]
8. Being the constant term:
[tex]\[
\left(\frac{\Delta H}{R} \right) = \frac{40790}{8.314}
\][/tex]
9. Now, putting it all together:
[tex]\[
T_2 = \frac{ \frac{\Delta H}{R } }{ \left(\frac{1}{373 \text{ K}} - \ln \left(\frac{79.8}{101.3 }\right)\right) }
\][/tex]
Substituting the values:
[tex]\[
T_2 = \frac{ \frac{40790}{8.314} }{ 0.2412438719463277 }
\][/tex]
Evaluating these calculations:
10. Finally, calculates:
[tex]\[
T_2 = 20337.023707393397 \text{ K}
\][/tex]
Thus, [tex]\( T_2 \)[/tex] is approximately [tex]\( 20337.02 \text{ K} \)[/tex].
