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3 years ago

7

The ratio of the initial to the final pressure of a given mass of gas is 1:1.5. Calculate the final volume of the gas if the ini

tial volume was 300cm^3 at the same temperature.

1 answer:

Paladinen [302]3 years ago

6 0

Answer:

Explanation:

Ratio of volume 1:1.5 = 300/v

1/1.5 = 300/v

Cross multiply

v = 1.5 × 300

V = 450cm^3

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Calculate average kinetic energy of one mole of gas at 517K

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In an ideal gas, there are no attractive forces between the gas molecules, and there is no rotation or vibration within the molecules. The kinetic energy of the translational motion of an ideal gas depends on its temperature. The formula for the kinetic energy of a gas defines the average kinetic energy per molecule. The kinetic energy is measured in Joules (J), and the temperature is measured in Kelvin (K).

K = average kinetic energy per molecule of gas (J)

kB = Boltzmann's constant ()

T = temperature (k)

Kinetic Energy of Gas Formula Questions:

1) Standard Temperature is defined to be . What is the average translational kinetic energy of a single molecule of an ideal gas at Standard Temperature?

Answer: The average translational kinetic energy of a molecule of an ideal gas can be found using the formula:

The average translational kinetic energy of a single molecule of an ideal gas is (Joules).

2) One mole (mol) of any substance consists of molecules (Avogadro's number). What is the translational kinetic energy of of an ideal gas at ?

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3 years ago

(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct

Zarrin [17]

Explanation:

The given data is as follows.

$\Lambda^{o}_{m}$ (NaCl) = $1.264 \times 10^{-2}$

$\Lambda^{o}_{m}$ (H-O=C-ONO) = $1.046 \times 10^{-2}$

$\Lambda^{o}_{m}$ (HCl) = $4.261 \times 10^{-2}$

Conductivity of monobasic acid is $5.07 \times 10^{-2} S m^{-1}$

Concentration = 0.01 $mol/dm^{3}$

Therefore, molar conductivity ( $\Lambda_{m}$ ) of monobasic acid is calculated as follows.

$\Lambda_{m} = \frac{conductivity}{concentration}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}$

= $5.07 \times 10^{-3} S m^{2} mol^{-1}$

Also, $\Lambda^{o}_{m}$ = $\Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}$

= $4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}$

= $4.043 \times 10^{-2} S m^{2} mol^{-1}$

Relation between degree of dissociation and molar conductivity is as follows.

$\alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}$

= 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

Putting the values into the above formula we get the following.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

= $\frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}$

= $0.017973 \times 10^{-2}$

= $1.7973 \times 10^{-4}$

Hence, the acid dissociation constant is $1.7973 \times 10^{-4}$ .

Also, relation between $pK_{a}$ and $K_{a}$ is as follows.

$pK_{a} = -log K_{a}$

= $-log (1.7973 \times 10^{-4})$

= 3.7454

Therefore, value of $pK_{a}$ is 3.7454.

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