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yawa3891 [41]
3 years ago
14

The vapor pressure of ethanol at 293 K is 5.95 kPa and at 336.5 K it is 53.3 kPa. Calculate the enthalpy of vaporization of etha

nol. (Note: any unit of pressure may be used, as long as you are doing P2/P1. Temp must be K)
Physics
1 answer:
denis-greek [22]3 years ago
4 0

Answer:

H=41.3kJmol^{-1}

Explanation:

The equation relating the the enthalphy, pressure and temperature is expressed as

ln(\frac{P_{2}}{P_{1}} )=\frac{H}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}} ) \\

Where P is the pressure, H is the enthalphy, and T is the temperature.

since the given values are

T_{1}=293k, \\T_{2}=336.5k\\P_{1}=5.95kPA\\p_{2}=53.3kPA\\and R=8.314J.K^{-1}mol_{-1}

if we insert values, we arrive at

ln(\frac{53.3}{5.95} )=\frac{H}{8.314}(\frac{1}{293}-\frac{1}{336.5} )\\2.19=\frac{H}{8.314}(0.00044)\\H=(2.19*8.314)/0.00044\\H=41,268.8Jmol^{-1}\\H=41.3kJmol^{-1}

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Answer:

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The equation of the system's heat exchange is given by

m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C

The final equilibrium temperature is 9.50022°C

4 0
3 years ago
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<h2>Answer:</h2>

-310J

<h2>Explanation:</h2>

The change in internal energy (ΔE) of a system is the sum of the heat (Q) and work (W) done on or by the system. i.e

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If heat is released by the system, Q is negative. Else it is positive.

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<em>In this case, the system is the balloon and;</em>

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<em>Substitute these values into equation (i) as follows;</em>

ΔE = -695 + 385

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Therefore, the change in internal energy is -310J

<em>PS: The negative value indicates that the system(balloon) has lost energy to its surrounding, thereby making the process exothermic.</em>

<em />

<em />

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What is the unit for electrical power?
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Answer:

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