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States of Matter Test - 54

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States of Matter Test - 54
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  • Question 1
    1 / -0
    Inelastic collisions occur in :
    I. Real gases
    II. Ideal gases
    III. Fusion reactions
    Which of the following is correct option about the above problem?
    Solution
    An inelastic collision is one in which part of the kinetic energy is changed to some other form of energy in the collision. Any macroscopic collision between objects will convert some of the kinetic energy into internal energy and other forms of energy,
    Such collisions occur in real gases and fusion reaction.
  • Question 2
    1 / -0
    To use Gay-Lussac's Law, which of the following needs to remain constant?
    Solution
    $$P \propto T $$
    At constant volume and number of moles of a gas, the pressure is directly proportional to the absolute temperature.
    Hence, option $$A$$ is correct.
  • Question 3
    1 / -0
    Avogadro's law shows the relationship between which two variables?
    Solution
    (A) : Volume and number of moles, : at constant pressure.
    Avogadro Law: At the same condition of temperature and pressure the volume of the gas is directly proportional to the number of the moles.
  • Question 4
    1 / -0
    Statement I : At STP, 22.4 liters of He will have the same volume as one mole of $$\displaystyle { H }_{ 2 }$$ (assume ideal gases).
    Statement II : One mole or 22.4 liters of any gas at STP will have the same mass.
    Solution
    Avogadro’s law states that under the same conditions of temperature and pressure, equal volumes of different gases contain an equal number of molecules. This empirical relation can be derived from the kinetic theory of gases under the assumption of a perfect (ideal) gas. The law is approximately valid for real gases at sufficiently low pressures and high temperatures.
    At STP, all gas have same volume for 1 mol of gas and that volume is always equal to the 22.4 L.  
  • Question 5
    1 / -0
    Dalton's law of partial pressure is applicable to which one of the following systems?
    Solution
    Hint: Dalton's law of partial pressure is applicable to non reacting gases 

    Explanation for Correct Answer:


    Dalton's law (also called Dalton's law of partial pressures) states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases.

    So, In order to follow the daltons law of partial pressure, the necessary condition is that gases would not react with themselves.

    In option (D), gases $$CO$$ and $$H_2$$ do not react with each other at normal temperature and for the reaction, it requires a higher temperature and pressure. due to which law is applicable here.

    Explanation for Incorrect Answer:

    In option (A), gases can react with each other due to which law is not applicable here.
    The reaction between $$NH_3$$ and $$HCl$$ is shown as;
    $$ NH_3 + HCl \rightarrow NH_4Cl$$

    In option (B)  also, gases can react with each other due to which law is not applicable here.
    The reaction between $$NO$$ and $$O_2$$ is shown as;
    $$ 2 NO + O_2 \rightarrow 2 NO_2$$

    In option (C) also, gases can react with each other due to which law is not applicable here.
    The reaction between $$H_2$$ and $$Cl_2$$ is shown as;
    $$ H_2 + Cl_2 \rightarrow 2 HCl$$


    Correct Option: $$D$$

  • Question 6
    1 / -0
    Two liquids $$X$$ and $$Y$$ form an ideal solution at $$300\ K$$, the vapour pressure of the solution containing $$1\ mole$$ of $$X$$ and $$3\ moles$$ of $$Y$$ is $$550$$ mm Hg. At the same temperature, if $$1\ mole$$ of $$Y$$ is further added to this solution, the vapour pressure of the solution increases by $$10\ mm\ Hg$$. Vapour pressure (in mm Hg) of $$X$$ and $$Y$$ in their pure states will be respectively :
    Solution
    $$P_{total}=P^o_A\cdot X_A+P^o_A\cdot X_B$$

    $$550=P^o_A+\times \dfrac{1}{4}+P^o_B\times \dfrac{3}{4}$$

    $$P^o_A+3P^o_B=2200$$                                 ...(i)

    When $$1\, mol$$ of $$y$$ is further added to the solution 
    $$560=P^o_A\times \dfrac{1}{5}+P^o_B\times \dfrac{4}{5}$$

    $$P^o_A+4P^o_B = 2800$$                             ...(ii)

    On subtraction, (ii) - (i)

    $$P^o_B=2800-2200=600$$

    On putting the value of $$P^o_B$$ in Eq. (i)

    $$P^o_A+3\times 600 = 2200$$

    $$P^o_A=2200-1800=400$$

    Hence, the correct option is $$C$$.
  • Question 7
    1 / -0
    Calculate the temperature at which the root mean square velocity of chlorine gas molecules is twice the root mean square velocity of carbon dioxide gas molecules at $$40^0C$$.
    Solution
    The expression for the root mean square velocity is $$\displaystyle u = \sqrt {\dfrac {3RT}{M}}$$

    For carbon dioxide gas
    $$\displaystyle u = \sqrt {\dfrac {3R \times (40+273)}{44}} $$

    $$\displaystyle u = \sqrt {3R \times7.114} $$  ...... (1)

    For chlorine gas
    $$\displaystyle u = \sqrt {\dfrac {3RT}{71}}$$  ..... (2)

    The root mean square velocity of chlorine gas molecules is twice The root mean square velocity of carbon dioxide gas molecules. 

    From (1) and (2)

    $$\displaystyle  \sqrt {\dfrac {3RT}{71}} =2 \times  \sqrt {3R \times7.114} $$

    $$\displaystyle  \dfrac {3RT}{71} =4 \times  3R \times7.114 $$

    $$\displaystyle  \dfrac {T}{71}$$ $$=4 \times  7.114 $$

    $$\displaystyle  T =71 \times 4 \times  7.114 $$

    $$T=2020 \: K \simeq 2000$$ K
  • Question 8
    1 / -0
    When $$100\ ml$$ of a $$O_{2} - O_{3}$$ mixture was passed through turpentine, there was reduction of volume by $$20\ mL$$. If $$100\ ml$$ of such a mixture is heated, what will be the increase in volume?
    Solution

    Given,

    Total volume of $${O_2} - {O_3}$$ mixture$$ = 100mL$$

    Reduction in volume$$ = 20mL$$

    We know that $${O_3}$$ gets absorbed in turpentine oil, therefore

    Volume of $${O_3}$$ that gets absorbed $$ = 20mL$$

    Volume of $${O_2} = (100 - 80)mL = 20mL$$

    On heating, the following reaction takes place between $${O_3}$$ and $${O_2}$$

    $${20_3} \rightleftharpoons 3{O_2}$$

    Since volume is proportional to moles, therefore

    $${20_3} \rightleftharpoons 3{O_2}$$

    $$2mol$$      $$3mol$$

    $$20mL$$        $$30mL$$

    Therefore, increase in volume will be $$ = (30 - 20)mL = 10mL$$

    Hence, percentage increase in volume will be = Increase in volume/Total volume *100

    $$ \Rightarrow \dfrac{{10mL}}{{100mL}} \times 100$$

    $$\Rightarrow 10$$%

    The correct answer will be option A.

  • Question 9
    1 / -0
    An ideal gas: 
    Solution

    No molecular forces are at work for ideal gases. This means that there is no attraction or repulsion between the particles. So there will be no inter molecular force of attraction is present between the molecules of ideal gases.

    Hence option A is correct.

  • Question 10
    1 / -0
    Which of the following shows behavior of binary liquid solution?
    Solution
    For binary liquid solution, plot of $$\displaystyle \dfrac { 1 }{ { P  }_{ total } }$$ vs $$\displaystyle { Y }_{ B }$$ is linear.

    $$\displaystyle \dfrac { { P }_{ B } }{ {P  }_{ total } } = { Y }_{ B }$$

    $$\displaystyle \dfrac { 1 }{ {P  }_{ total } }  \propto { Y }_{ B }$$
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