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2 years ago

• How did your experimental absolute zero value compare to the accepted value?

Chemistry

1 answer:

Murrr4er [49]2 years ago

5 0

The experimental absolute zero value is less when compared to the accepted value of absolute zero.

<h3>What is absolute zero?</h3>

Absolute zero is defined as the temperature in which the lowest energy possible is attained in a thermodynamic system.

Absolute zero temperature has an accepted values of 0 Kelvin or -273.15 degrees Celsius.

At absolute zero, it is assumed that the volume of an ideal gas becomes zero. However, it has not been possible to cool any gas to absolute zero.

Based on the graph of temperature against volume of gases, the experimental absolute zero extrapolated from the graph where volume of the gases becomes zero is -285 degrees Celsius.

Therefore, the experimental absolute zero value is less when compared to the accepted value.

Learn more about absolute zero at: brainly.com/question/1191114

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The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment

astraxan [27]

Answer:

The activation energy is $=8.1\,kcal\,mol^{-1}$

Explanation:

The gas phase reaction is as follows.

$A \rightarrow B+C$

The rate law of the reaction is as follows.

$-r_{A}=kC_{A}$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = $10dm^{3}$

Temperature "T" = 300 K

Volumetric flow rate of the reaction $v_{o}=5dm^{3}s$

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

$V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]$

Rearrange the formula is as follows.

$k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_{o}}$ is 1.

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}$

Therefore, the rate constant in case of the plug flow reacor at 300K is $1.2s^{-1}$

The rate constant in case of the CSTR can be calculated by using the formula.

$\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)$

The feed has 50% A and 50% inerts.

Hence, the mole fraction of A in feed $y_{A_{o}}$ is 0.5

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacted.

$=0.5(2-1)=0.5$

Substitute the all values in formula (2)

$\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^{-1}$

The activation energy of the reaction can be calculated by using formula

$k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]$

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

$E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}$

Substitute the all values.

$=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}$

$=8.1\,kcal\,mol^{-1}$

Therefore, the activation energy is $=8.1\,kcal\,mol^{-1}$

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Answer:

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2 years ago

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nekit [7.7K]

Oxygen (O) = 42,129%

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3 years ago

A gas-filled weather balloon with a volume of 65.0 L is released at sea level where conditions are745 torr and 25 ºC. The balloo

Blababa [14]

Answer:

It will reach to its maximum volume.

Explanation:

Using Ideal gas equation for same mole of gas as :-

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

Given ,

V₁ = 65.0 mL

V₂ = ?

P₁ = 745 torr

The conversion of P(torr) to P(atm) is shown below:

$P(torr)=\frac {1}{760}\times P(atm)$

So,

Pressure = 745 / 760 atm = 0.9803 atm

P₁ = 0.9803 atm

P₂ = 0.066 atm

T₁ = 25 ºC

T₂ = -5 ºC

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (25 + 273.15) K = 298.15 K

T₂ = (-5 + 273.15) K = 268.15 K

Using above equation as:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac{{0.9803}\times {65.0}}{298.15}=\frac{{0.066}\times {V_2}}{268.15}$

Solving for V₂ as:-

$V_2=\frac{0.9803\times \:65.0\times 268.15}{298.15\times 0.066}$

$V_2=\frac{17086.38392}{19.6779}$

V₂ = 868 L

Given that:- V max = 835 L

<u>Thus, it will reach to its maximum volume.</u>

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3 years ago