A certain substance has a heat of vaporization of 62.08 kJ/mol. At what Kelvin temperature will the vapor pressure be 4.50 times higher than it was at 361 K?
I've gotten it to the point of:
ln(4.50)=7466.47423176378616 K (1/361 - 1/T2)
Solving for T2 but I don't know where to go from here!
To solve for T2, you need to isolate it in the equation. Let's break down the steps to solve the equation:
1. Start with the equation you have: ln(4.50) = 7466.47423176378616 K (1/361 - 1/T2)
2. Simplify the equation by multiplying both sides by the denominator of the second term on the right side (361T2):
ln(4.50) * 361T2 = 7466.47423176378616 K (361T2 / 361 - 361 / T2)
3. Distribute the terms on the right side:
ln(4.50) * 361T2 = 7466.47423176378616 K (T2 - 361)
4. Expand further:
ln(4.50) * 361T2 = 7466.47423176378616 K * T2 - 7466.47423176378616 K * 361
5. Simplify the equation by moving all terms involving T2 to one side and the constant terms to the other side:
ln(4.50) * 361T2 - 7466.47423176378616 K * T2 = -7466.47423176378616 K * 361
6. Factor out T2 from the left side:
T2 (ln(4.50) * 361 - 7466.47423176378616 K) = -7466.47423176378616 K * 361
7. Finally, solve for T2 by dividing both sides by (ln(4.50) * 361 - 7466.47423176378616 K):
T2 = (-7466.47423176378616 K * 361) / (ln(4.50) * 361 - 7466.47423176378616 K)
Using a calculator, you can substitute the values and calculate the solution:
T2 ≈ 374.89 K
Therefore, at approximately 374.89 K, the vapor pressure will be 4.50 times higher than it was at 361 K.