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States of Matter Test - 52

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States of Matter Test - 52
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  • Question 1
    1 / -0
    The compound A undergoes reaction as shown above.

    Find out the reduction product of $$'A'$$.

  • Question 2
    1 / -0
    A mixture of hydrogen and helium is prepared such that the number of wall collisions per unit time by molecules of each gas is the same. Which gas has the higher concentration?
    Solution
    The number of wall collisions per unit time by molecules of each gas is the same. Hence, the pressure is the same.

    $$\displaystyle P=CRT $$

    Since pressure is directly proportional to concentration, the two gases have equal concentration.

    Hence, the correct answer is option $$\text{C}$$.
  • Question 3
    1 / -0

    Directions For Questions

    Collision cross-section is an area of an imaginary sphere of radius $$\sigma $$ around the molecule within which the center of another molecule cannot penetrate. The volume swept by a single molecule in unit time is,
    $$V=(\pi \sigma ^{2}) \bar{u}$$, where $$\bar {u}$$ is the average speed.
    If $$N^{*}$$ is the number of molecules per unit volume, then the number of molecules within the volume $$V$$ is 
    $$N=VN^{*}=(\pi \sigma ^{2} \bar{u})N^{*}$$
    Hence, the number of collision made by a single molecule in unit time will be,
    $$Z=N=(\pi \sigma ^{2} \bar{u})N^{*}$$
    In order to account for the movements of all molecules, we must consider the average velocity along the line of centres of two colliding molecules instead of the average velocity of a single molecule. If it is assumed that, on an average, molecules collide while approaching each other perpendicularly, then the average velocity along their centres is $$\sqrt{2} \bar {u}$$ as shown below,
    Number of collision made by a single molecule with other molecules per unit time is given by,
    $$Z_{1}=\pi \sigma ^{2} (\bar {u_{rel}}) N^{*}=\sqrt {2}\pi \sigma ^{2} \bar {u} N^{*}$$
    The total number of bimolecular collisions $$Z_{11}$$ per unit volume per unit time is given by,
    $$Z_{11}=\displaystyle \frac{1}{2}(Z_{1}N^{*})$$ or $$Z_{11}=\displaystyle \frac{1}{2}(\sqrt{2}\pi \sigma ^{2}\bar{u}N^{*})N^{*}=\displaystyle \frac{1}{\sqrt{2}}\pi \sigma ^{2}\bar{u}N^{*2}$$
    If the collision involve two unlike molecules then the number of collisions $$Z_{12}$$ per unit volume per unit time is given as,
    $$Z_{12}=\pi \sigma _{12}^{2}\left ( \sqrt{\displaystyle \frac{8kT}{\pi \mu }} \right )N_{1}N_{2}$$
    where $$N_{1}$$ and $$N_{2}$$ are the number of molecules per unit volume of the two types of molecules, $$\sigma _{12}$$ is the average diameter of the two molecules and $$\mu$$ is the reduced mass. The mean free path is the average distance travelled by a molecule between two successive collisions. We can express it as follows;
    $$\lambda =\displaystyle \frac{Average\: distance\: travelled\: per\: unit\: time}{No.\: of\: collisions\: made\: by\: a\: single\: molecule\: per\: unit\: time}=\displaystyle \frac{\bar{u}}{Z_{1}}$$
    or $$\lambda =\displaystyle \frac{\bar {u}}{\sqrt{2}\pi \sigma ^{2}\bar {u}N^{*}}\Rightarrow \displaystyle \frac{1}{\sqrt{2}\pi \sigma ^{2}N^{*}}$$
    Three ideal gas samples in separate equal volume containers are taken and following data are recorded :


    PressureTemperatureMean free pathsMol.wt
    Gas A1 atm1600 K0.16 nm20
    Gas B2 atm200 K0.16 nm40
    Gas C4 atm400 K0.04 nm80

    ...view full instructions

    Calculate the ratio of collision frequencies $$(Z_{11}) $$ for the gases $$A,\ B,\ C$$ respectively.

    Solution
    Let $$\sigma _{A}, \sigma_{B}$$ and $$\sigma_{C}$$ be collision diameters of $$A,B$$ and $$C$$.

    $$\lambda=\displaystyle \frac{kT}{\sqrt{2} \pi \sigma^{2}P}$$

    $$ i.e, \sigma \propto \sqrt {\displaystyle \frac{T}{\lambda P}}$$

    $$\therefore\      \sigma_{A}:\sigma_{B}:\sigma_{C}$$

    $$\Rightarrow\sqrt {\displaystyle \frac{1600}{0.16 \times 1}}:\sqrt {\displaystyle \frac{200}{0.16 \times 2}}: \sqrt {\displaystyle \frac{400}{0.04 \times 4}}$$

    $$=4:1:2$$

    The ratio of collision frequencies is :
    $$Z_{11} \propto \displaystyle \frac{\sigma ^{2}P^{2}}{\sqrt{mT^{3/2}}}$$

    $$\Rightarrow A:B:C=\displaystyle \frac{16 \times 1}{\sqrt{20} \times (1600)^{3/2}}:\displaystyle \frac{1 \times 4}{\sqrt{40} \times (200)^{3/2}}:\displaystyle \frac{4 \times 16}{\sqrt{80} \times (400)^{3/2}}$$

    $$=1:4:16$$
    Hence, option $$(C)$$ is correct.
  • Question 4
    1 / -0
    Calculate the mean free path in $$\displaystyle CO_{2}$$ at $$\displaystyle 27^{\circ}C$$ and a pressure of $$\displaystyle 10^{-9}$$ bar. Molecular diameter of $$CO_2$$ is $$500$$ pm.
    [ Given : R $$\displaystyle =\frac{25}{3}$$ $$\displaystyle J.mol^{-1}K^{-1}, \sqrt{2}=1.4, \pi =\frac{22}{7}, N_{A}=6\times 10^{23}$$]
    Solution
    The number of molecules per unit volume $$\displaystyle N = \frac {n}{V} \times N_A=   \frac {P}{RT} \times N_A=   \frac {10^{-9}}{1.01325 \times 0.0821 \times 300} \times 6.023 \times 10^{23} $$

    $$= 2.4 \times 10^{13} \: \text{molecules/dm}^3 = 2.4 \times 10^{10} \: \text{molecules/cm}^3$$

    The mean free path is $$\displaystyle \frac {1}{\sqrt {2} \times \pi \times \sigma ^2 \times N} =   \frac {1}{1.414 \times 3.1416 \times (500 \times 10^{-10})^2 \times 2.4 \times 10^{+10}} = 3.788 \times 10^3 \: cm $$
  • Question 5
    1 / -0
    For an ideal gas, number of mol per litre in terms of its pressure p, temperature T and gas constant R is:
    Solution

  • Question 6
    1 / -0
    In the reaction $$N_{2}+3H_{2}\rightarrow 2NH_{3} $$, the ratio by volume of $$N_{2},\ H_{2} \:$$ and$$\: NH_{3}$$ is $$1 : 3 : 2$$. 

    This illustrates the law of:
    Solution
    In the reaction, $$N_{2}+3H_{2}\rightarrow 2NH_{3} $$, the ratio by volume of $$N_{2},\ H_{2} $$ and $$ NH_{3}$$ is $$1 : 3 : 2$$. This illustrates the law of Gaseous volumes or Gay Lussac's law of combining volumes of gases.

    According to this law, when gases react together to produce gaseous products, the volume of reactants and products bear a simple whole-number ratio with each other, provided volumes are measured at the same temperature and pressure.

    So, the correct option is $$D$$.
  • Question 7
    1 / -0
    A balloon containing 1 mole air at 1 $$atm$$ initially is filled further with air till pressure increases to 4 $$atm$$. The initial diameter of the balloon is 1 $$m$$ and the pressure at each stage is proportion to diameter of the balloon. How many no.of moles of air added to change the pressure from 1 $$atm$$ to 4 $$atm$$.
    Solution
    Pis directly proportional to d; P=kd  
    and k=1atm/1metre
    $$PV=nRT;kd(\dfrac{1}{6}\pi d^3)=nRT$$
    $$\dfrac{d_1^4}{d_2^4}=\dfrac{n_1}{n_2}$$
    $$\dfrac{1}{4^4}=\dfrac{n_1}{n_2}$$
    $$n_2=256$$
    No.of moles added=256−1=255.
  • Question 8
    1 / -0
    "All gases have the same number of moles in the same volume at constant temperature and pressure:. This statments belongs to :
    Solution
    Avogadro's law (sometimes referred to as Avogadro's hypothesis or Avogadro's principle) is an experimental gas law relating volume of a gas to the amount of substance of gas present.
    Avogadro's law states that, "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules".
    For a given mass of an ideal gas, the volume and amount (moles) of the gas are directly proportional if the temperature and pressure are constant.
  • Question 9
    1 / -0
    Ideal gas has a volume of 10 litres at $$20^oC$$ and a pressure of 750 $$mm$$ $$Hg$$. Find the expressions which is needed  to determine the volume of the same amount of gas at $$STP$$?
    Solution
    Initial volume $$V_1  = 10 \:L$$
    Final volume $$V_2  = ?$$

    Initial pressure $$P_1 = 750 \:mm \:Hg$$
    Final pressure $$P_2 = 760 \:mm \:Hg$$

    Initial temperature $$T_1 =20^oC =  293 \:K$$
    Final temperature $$T_2 = 273 \:K$$
     
    $$\displaystyle \dfrac {P_1V_1}{T_1}=\dfrac {P_2V_2}{T_2} \\ \dfrac {750 \times 10}{293}=\dfrac {760 \times V2}{273} \\ V_2 = 10 \times \dfrac {750}{760} \times \dfrac {273}{293}$$

    Hence, the option (E) is the correct answer.
  • Question 10
    1 / -0
    3000 cc of oxygen was burnt with 600 cc of ethane($$C_2H_6$$). Calculate the volume of unused oxygen.
    Solution
    The reaction is as follows:
    $$C_2H_6+\dfrac72O_2\longrightarrow2CO_2+3H_2O$$
    600 cc. of ethane reacts with $$600\times\dfrac72$$, i.e., 2100 cc volume of oxygen. 
    Remaining volume of $$O_2$$ is $$3000-2100=900\  cc.$$
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