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dezoksy [38]
3 years ago
14

Consider two flasks at 25 degree Celsius, one contains an ideal gas and the other contains the real gas SO3. Which statement reg

arding these gases is true?
A. As the temperature approaches 0 K, the volume of the ideal gas will be larger than the volume of SO3 because ideal gases lack inter-molecular forces.
B. As the temperature is decreased, the ideal gas will liquefy first because ideal gases have larger values of the van der Waals coefficient b.
C. As the temperature approaches 0 K, the volume of the ideal gas will be smaller than the volume of SO3 because ideal gases have larger values of the van der Waals coefficient a.
D. After the temperature has been lowered about 100 K, the pressure of the ideal gas will be smaller than the pressure of SO3 because the van der Waals coefficient a for SO3 is large.
E. After the temperature is increased about 100 K, the pressure of the ideal gas will be smaller than the pressure of SO3 because the van der Waals coefficient a for SO3 is lar
Chemistry
1 answer:
Strike441 [17]3 years ago
3 0

Answer: Option (A) is the correct answer.

Explanation:

In real gases, there exists force of attraction between the molecules at low temperature and high pressure. This is because at low temperature there occurs a decrease in kinetic energy of gas molecules and high pressure causes the molecules to come closer to each other.

As a result, forces of attraction increases as molecules come closer to each other and therefore, gases deviate from an ideal gas behavior.

And, at low pressure and high temperature there exists no force of attraction or repulsion between the molecules of a gas because they have high kinetic energy. Hence, gases behave ideally at these conditions.

Thus, we can conclude that the statement as the temperature approaches 0 K, the volume of the ideal gas will be larger than the volume of SO_{3} because ideal gases lack inter-molecular forces, is true.

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Hello,
nikitadnepr [17]

Answer:

a)\ 2FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3\\b)\ Decomposition\ Reaction

Explanation:

<em>Ferrous Sulphate</em>(FeSO4)<em> is generally found as Lime-Green Crystals. On heating, these crystals almost immediately turn white-yellow. They then, break down to produce an anhydrous mixture of Sulphur Trioxide </em>(SO_3)<em>, Sulphur Dioxide </em>(SO_2)<em>  as well as Ferric Oxide </em>(Fe_2O_3)<em>.</em>

<em>We can hence, frame a skeletal equation of this reaction and try to balance it.</em>

<em>Hence,</em>

FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3

<em>Now,</em>

<em>a)In order to balance it through the 'Hit &Trial Method', we'll follow a series of </em><em>steps</em><em>:</em>

<em>1. First, lets compare the number of  Fe (Iron) atoms on the RHS and LHS. We find that, the no. of Fe Atoms on the RHS is twice the number of Fe Atoms on the LHS. We hence, add a co-effecient 2 beside </em>FeSO_4.

<em>2. Now, Iron atoms, Sulphur Atoms and Oxygen atoms occur 2, 2, 8 respectively on both the sides:</em>

<em> Hence, As all the other elements as well as iron, balance, we've arrived upon our Balanced Equation :</em>

<em> </em>2FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3

<em>b) We know that, decomposition reactions are [generally] endothermic reactions in which Large Compounds </em><em>decompose </em><em>into smaller elements and compounds. Here, as Ferrous Sulphate </em><em>decomposes </em><em>into Sulphur Dioxide, Sulphur Trioxide and Ferric Oxide, the reaction that occurs here is </em><em>Decomposition Reaction.</em>

7 0
3 years ago
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