Question

A particular compound has an enthalpy of vaporization of 28900 J/mol. At 278 K it has a vapor pressure of 103 mmHg. What is its vapor pressure at 309 K? (R = 8.31 J/(K· mol))a. 29.4mmhgb. 99.5mmhgc. 107mmhgd. 194mmhge. 361mmhg

Answer 1

`E\text{ : 361 mmHg}`

Step-by-step explanation:

Here, we want to get the vapour pressure of the compound at the given temperature

We can use the Clausius-Clapeyron equation for this

Mathematically, we have this as:

`\ln ((P_1)/(P_2))\text{ = -}(\Delta H)/(R)((1)/(T_1)-(1)/(T_2))`

So,let us identify the values:

P1 = 103 mmHg

P2 = ?

ΔH = 28,900

R = 8.31

T1 = 278 K

T2 = 309 K

We now proceed to substitute these values into the equation above as follows:

`\begin{gathered} \ln ((103)/(P_2))\text{ = -}(28900)/(8.31)((1)/(278)-(1)/(309)) \\ \\ \ln ((103)/(P_2))\text{ = -3477.74 (}(31)/(85902)) \\ \\ \ln ((103)/(P_2))\text{ = -1.255} \\ \\ e^(-1.255)\text{ = }(103)/(P_2) \\ P_2\text{ = 361 mmHg} \end{gathered}`