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SSSSS [86.1K]
3 years ago
11

A certain substance has a heat of vaporization of 64.08 kJ/mol. At what Kelvin temperature will the vapor pressure be 6.50 times

higher than it was at 355 K?
Chemistry
1 answer:
ValentinkaMS [17]3 years ago
3 0

Answer:

T_2 =388.50K

Explanation:

Given

p_1 = ?? -- Initial vapour pressure

p_2 = 6.50p_1 --- Final vapour pressure

T_1 = 355K ---- Initial temperature

T_2 = ?? --- Final temperature

\triangle _{vap}H = 64.08kJmol --- Enthalpy of vaporization

Required

Calculate T2

To do this, we make use of Clausius-Clapeyron equation.

Which states that:

ln(\frac{p_2}{p_1}) = \frac{\triangle _{vap}H}{R}(\frac{1}{T_1} - \frac{1}{T_2})

Where:

R = 8.314 J.K^{-1}mol^{-1} --- Universal Gas constant

\triangle _{vap}H = 64.08kJmol

\triangle _{vap}H = 64080\ kJmol

ln(\frac{p_2}{p_1}) = \frac{\triangle _{vap}H}{R}(\frac{1}{T_1} - \frac{1}{T_2}) becomes

ln(\frac{6.50p_1}{p_1}) = \frac{64080}{8.314}(\frac{1}{355} - \frac{1}{T_2})

ln(\frac{6.50p_1}{p_1}) = 7707.48(\frac{1}{355} - \frac{1}{T_2})

ln(6.50) = 7707.48(\frac{1}{355} - \frac{1}{T_2})

1.872 = 7707.48(\frac{1}{355} - \frac{1}{T_2})

Take LCM

1.872 = 7707.48(\frac{T_2 - 355}{355T_2})

1.872 = 7707.48*\frac{T_2 - 355}{355T_2}

1.872 = \frac{7707.48*(T_2 - 355)}{355T_2}

Cross Multiply

355T_2*1.872 = 7707.48*(T_2 - 355)

664.56T_2 = 7707.48T_2 - 2736155.4

Collect Like Terms

664.56T_2 - 7707.48T_2 =- 2736155.4

-7042.92T_2 =- 2736155.4

Make T2 the subject

T_2 =\frac{- 2736155.4}{-7042.92}

T_2 =388.497299416

T_2 =388.50K

<em>The final temperature is 388.50K</em>

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