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ivanzaharov [21]
3 years ago
12

How many moles of a gas sample are in a 5.0 L container at 205 K and 340 kPa?

Chemistry
1 answer:
Mazyrski [523]3 years ago
3 0

Answer:

1.0 mole

Explanation:

From the question given above, the following data were obtained:

Volume (V) = 5 L

Temperature (T) = 205 K

Pressure (P) = 340 KPa

Gas constant (R) = 8.31 KPa.L/Kmol

Number of mole (n) =?

Using the ideal gas equation, the number of mole of the gas in the container can be obtained as follow:

PV = nRT

340 × 5 = n × 8.31 × 205

1700 = n × 1703.55

Divide both side by 1703.55

n = 1700 / 1703.55

n = 1.0 mole

Thus, the number of mole of the gas in the container is 1.0 mole

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3 years ago
Calculate the number of molecules?
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87.33 X 10^15 molecules
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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
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Explanation :

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

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We know that,

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

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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 azide ion, n−3, is a symmetrical ion, all of whose contributing structures have formal charges. draw three important contrib
stich3 [128]

Explanation:

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monitta

Answer: (2) releases 2260 J/g of heat energy

Explanation:

Latent heat of vaporization is the amount of heat required to convert 1 mole of liquid to gas at atmospheric pressure.

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The temperature does not change during this process, so heat released goes into changing the state of the substance, thus it is called latent which means hidden. The energy released in this process is same in magnitude as latent heat of vaporization. The heat of condensation of water vapour is about 2,260 J/g.

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