Solve Ethanol ΔHvap as a Function of Temp at Constant Pressure

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In summary, to find ΔHvap as a function of temperature, we can use the Clapyeron equation and integrate it with respect to temperature, using the ideal gas law and the definition of enthalpy change. This will give us the desired equation for ΔHvap.
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Homework Statement


The Clausius-Clapyeron equation ln(p2/p1) = -ΔHvap/R (1/t2 - 1/t1) is derived assuming that ΔHvap is independent of temperature.
Find ΔHvap as a function of temperature for ethanol at constant external pressure.

Homework Equations



these are not relevant equations but the equations that i tried solving using-
dU = TdS − pdV

The Attempt at a Solution


i tried starting with the equation for dh and integrating it. i also tried to derive using the clapyeron equation dp=(ΔH/tΔv) dt but i keep getting stuck. the other half of the question asks to derive the clausius clapyeron using the ΔH of this question so I am skeptic about the way I have approached this question.
 

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To find ΔHvap as a function of temperature, we can use the following steps:

1. Start with the definition of enthalpy: dH = TdS + Vdp

2. Use the Clapyeron equation to write dp = (ΔHvap/TΔv)dt

3. Substitute this expression for dp into the definition of enthalpy, and rearrange the terms to get:
dH = TdS + (ΔHvap/TΔv)dtV

4. Use the definition of entropy, dS = dq/T, to rewrite dS in terms of heat (dq) and temperature (T).

5. Substitute this expression for dS into the equation for dH, and rearrange the terms to get:
dH = dq + (ΔHvap/TΔv)dtV

6. Since the process is at constant external pressure, dq = dHvap. Substitute this into the equation to get:
dH = dHvap + (ΔHvap/TΔv)dtV

7. Integrate both sides of the equation with respect to temperature (t), from t1 to t2:
∫dH = ∫dHvap + (ΔHvap/TΔv)∫dtV

8. Simplify the integral on the right side to get:
H2 - H1 = ΔHvap + (ΔHvap/TΔv)(V2 - V1)

9. Use the ideal gas law, pV = nRT, to rewrite V in terms of pressure (p) and temperature (T).

10. Substitute this expression for V into the equation and rearrange the terms to get:
H2 - H1 = ΔHvap + (ΔHvap/TΔv)(nR/p)(T2 - T1)

11. Finally, use the definition of enthalpy change (ΔH = H2 - H1) to get:
ΔHvap = ΔH - (ΔHvap/TΔv)(nR/p)(T2 - T1)

This is the desired equation for ΔHvap as a function of temperature for ethanol at constant external pressure.
 

Related to Solve Ethanol ΔHvap as a Function of Temp at Constant Pressure

1. What is ΔHvap?

ΔHvap is the enthalpy of vaporization, also known as the heat of vaporization. It is the amount of heat required to convert a substance from liquid to gas at a constant pressure.

2. How does ethanol ΔHvap change with temperature?

As temperature increases, the ΔHvap of ethanol also increases. This is because higher temperatures provide more energy to break the intermolecular forces holding the liquid molecules together, making it easier for them to escape into the gas phase.

3. Why is pressure kept constant in this calculation?

Pressure is kept constant because enthalpy is a state function, meaning it only depends on the initial and final states of a substance. By keeping pressure constant, we can accurately measure the change in enthalpy without the effects of pressure changes.

4. How is ΔHvap determined experimentally?

ΔHvap can be determined experimentally by measuring the temperature and pressure of a liquid as it is heated and turned into a gas. The change in temperature and pressure can then be used to calculate the ΔHvap using the Clausius-Clapeyron equation.

5. What factors can affect the accuracy of the calculated ΔHvap?

The accuracy of the calculated ΔHvap can be affected by factors such as experimental errors, impurities in the substance, and variations in pressure and temperature. It is important to control these factors as much as possible to obtain an accurate value for ΔHvap.

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