Calculate the decrease in temperarure when 2.00 L at 20.0 degrees celsius is compressed to 1.00 L.

1 answer:

3 0

I suspect that the pressure of this change is constant therefore

The equation is used from the combined gas law. (When pressure is constant both P's will cancel out P/P = 1)
V/T = V/T
Initial Change

Initially we have 2L at 20 degress what temperature will be at 1L.

2/20 = 1/T
0.1 = 1/T
0.1T = 1
T = 1/0.1
T = 10 degress celsius.

Hope this helps if you won't be able to understand what is the combined gas law just tell me :).

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vladimir2022 [97]

The correct answer is b

7 0

2 years ago

The estimated heat of vaporization of diethyl ether using the Chen's rule is A. 29.7 KJ/mol B. 33.5 KJ/mol C. 26.4 KJ/mol D. 36.

Brums [2.3K]

Answer:

C. 26.4 kJ/mol

Explanation:

The Chen's rule for the calculation of heat of vaporization is shown below:

$\Delta H_v=RT_b\left [ \frac{3.974\left ( \frac{T_b}{T_c} \right )-3.958+1.555lnP_c}{1.07-\left ( \frac{T_b}{T_c} \right )} \right ]$

Where,

$\Delta H_v$ is the Heat of vaoprization (J/mol)

$T_b$ is the normal boiling point of the gas (K)

$T_c$ is the Critical temperature of the gas (K)

$P_c$ is the Critical pressure of the gas (bar)

R is the gas constant (8.314 J/Kmol)

For diethyl ether:

$T_b=307.4\ K$

$T_c=466.7\ K$

$P_c=36.4\ bar$

Applying the above equation to find heat of vaporization as:

$\Delta H_v=8.314\times307.4 \left [ \frac{3.974\left ( \frac{307.4}{466.7} \right )-3.958+1.555ln36.4}{1.07-\left ( \frac{307.4}{466.7} \right )} \right ]$