(a)
Interpretation:
The temperature dependence of the vapor pressure of phosphoryl chloride difluoride is given, ln p against 1/T has to be plotted.
Concept introduction:
The relationship to plot is In P versus 1/T is given below.
In P=−ΔHovapRT+ ΔSovap R
Slope =−ΔHovapR,Intercept = ΔSovapR
(a)

Explanation of Solution
Given,
Temperature, T/K | 190 | 228 | 250 | 273 |
Vapour pressure, P/Torr | 3.2 | 68 | 240 | 672 |
Unit conversion is given below,
1 Torrr = 0.00131579 atm(1) 3.22 Torr = 0.0042 atmln p = ln (0.0042 atm) = −5.47(2) 68 Torr = 0.089 atmln p = ln (0.089 atm) = −2.41(3) 240 Torr = 0.316 atmln p = ln (0.316 atm) = −1.15(4) 672 Torr = 0.884 atmln p = ln (0.884 atm) = −0.123
(1)Temperature (T)=190,(1T) = 0.00526 (2) Temperature (T)=228,(1T) = 0.00438(3) Temperature (T)=250,(1T) = 0.004(4) Temperature (T)=273,(1T) = 0.00366
Therefore,
Temp. (K) | T−1 (K−1) | V.P (Torr) | V.P (atm) | lnP |
190 | 0.00526 | 3.22 | 0.0042 atm | −5.47 |
228 | 0.00438 | 68 | 0.089 atm | −2.41 |
250 | 0.004 | 240 | 0.316 atm | −1.15 |
273 | 0.00366 | 672 | 0.884 atm | −0.123 |
The ln p against 1/T plot is given below,
Figure 1
From the curve, slope is given below,
y = mx + cy = (−3358.714)x + 12.247
(b)
Interpretation:
The enthalpy of the vaporization has to be calculated.
Concept introduction:
Refer to part (a)
(b)

Explanation of Solution
The relationship to plot is In P versus 1/T, this gives a straight line.
From the curve, slope is given below,
y = mx + cy = (−3358.714)x + 12.247
Slope =−ΔHovapR,− 3359 =−ΔHovap(8.314J⋅K−1⋅mol−1)ΔHovap = 28 kJ⋅mol−1
The enthalpy of the vaporization is 28 kJ⋅mol−1
(c)
Interpretation:
The entropy of the vaporization has to be calculated.
Concept introduction:
Refer to part (a)
(c)

Explanation of Solution
The relationship to plot is In P versus 1/T, this gives a straight line.
From the curve, slope is given below,
y = mx + cy = (−3358.714)x + 12.247
Intercept = ΔSovapR12.25 =ΔSovap(8.314J⋅K−1⋅mol−1)ΔSovap = 101.8 J⋅K−1⋅mol−1
The entropy of the vaporization is 101.8 J⋅K−1⋅mol−1.
(d)
Interpretation:
The normal boiling point of phosphoryl chloride difluoride has to be calculated.
Concept introduction:
Refer to part (a)
(d)

Explanation of Solution
The normal boiling point of phosphoryl chloride difluoride is calculated when the vapor pressure is 1 atm and ln (1) = 0.
Therefore, it will happen when.
T−1 = 0.0036T = 2.7 ×102 K
The normal boiling point of phosphoryl chloride difluoride is 2.7 ×102 K
(e)
Interpretation:
If the pressure is 15 torr, temperature of phosphoryl chloride difluoride has to be calculated.
Concept introduction:
Refer to part (a)
(e)

Explanation of Solution
The temperature of phosphoryl chloride difluoride value is calculated as follows,
In P=−ΔHovapRT+ ΔSovap RIn (15 Torr760 Torr)=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )T+(100 J⋅K−1⋅mol−1)(8.314J⋅K−1⋅mol−1 )−3.93=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )T+12.028−15.96=−(28000 J⋅mol−1)(8.314J⋅K−1⋅mol−1 )TT = 211 K
The temperature of phosphoryl chloride difluoride value is 211 K
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