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3 years ago

How many moles of gas are present in 1.13 L of gas at 2.09 atm and 291 K?

Chemistry

1 answer:

raketka [301]3 years ago

3 0

<h2><u>Answer:</u></h2>

n = 0.0989 moles

<h2><u>Explanation:</u></h2>

n = PV / RT

P = 2.09atm

V = 1.13L

R = 0.08206

T = 291K

Plug the numbers in the equation.

n = (2.09atm)(1.13L) / (0.08206)(291K)

n = 0.0989 moles

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Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h

bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

$\Delta S=-n\times C_v\ln \frac{T_2}{T_1}$

We know that,

The relation between the $C_p\text{ and }C_v$ for an ideal gas are :

$C_p-C_v=R$

As we are given :

$C_p=28.253J/K.mole$

$28.253J/K.mole-C_v=8.314J/K.mole$

$C_v=19.939J/K.mole$

Now we have to calculate the entropy change of the gas.

$\Delta S=-n\times C_v\ln \frac{T_2}{T_1}$

$\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J$

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. $(w=-pdV)$

(C) Heat during the process will be,

$q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J$

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

7 0

3 years ago

The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3

9966 [12]

Answer:

The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Where,

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314×10⁻³ kJ / K mol

$k_2=2.3\times 10^8$

$T_2=280\ ^0C$

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

T(K) = T( °C) + 273.15

So,

T = (280 + 273.15) K = 553.15 K

$T_2=553.15\ K$

$k_1=4.8\times 10^8$

$T_1=376\ ^0C$

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

T(K) = T( °C) + 273.15

So,

T = (376 + 273.15) K = 649.15 K

$T_1=649.15\ K$

So,

$\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)$

$E_a=-\frac{10^{-3}\times \:8.314\ln \left(\frac{10^8\times \:4.8}{10^8\times \:2.3}\right)}{-\frac{96}{359077.3225}}\ kJ/mol$

$E_a=-\frac{\frac{8.314\ln \left(\frac{4.8}{2.3}\right)}{1000}}{-\frac{96}{359077.3225}}\ kJ/mol$

$E_a=22.87\ kJ/mol$

<u>The activation energy for this reaction = 23 kJ/mol.</u>

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