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States of Matter Test - 13

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States of Matter Test - 13
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  • Question 1
    1 / -0
    The pressure of a gas is due to:
    Solution
    When the molecules hit the walls of the container, they exert a force on the container. This force appears as pressure on the walls of the container.

    The collision of molecules with each other will not generate pressure on the walls of the container.

    Neither will the random motion of gas molecules generate pressure.

    Hence, the correct answer is option $$\text{D}$$.
  • Question 2
    1 / -0
    Gram molar volume for a gas is always considered at ___ conditions.
    Solution
    Since volume is a dependent variable and it depends on pressure and temperature always so temperature and pressure should be specified when volume of gas is considered. For this certain standard conditions are taken as 273 K and 1 atm pressure.
    And at STP, molar volume is 22.4 litre.
  • Question 3
    1 / -0
    The volume occupied by 0.01 moles of helium gas at STP is:
    Solution
    As 1 mole of a gaseous substance occupies 22.4 litre ($$22.4 l =$$ gram molar volume)

    Volume occupied $$=GMV \times 0.01$$( gram molar volume )

    Hence, volume occupied $$= 22.4\times0.01$$
                                               $$= 0.224l$$
  • Question 4
    1 / -0
    In a 2.5 $$L$$ flask at $$27^{\circ}C$$ temperature, the pressure of a gas was found to be 8 $$atm$$. If $$2.41 \times 10^{23}$$ molecules of the same gas are introduced into the container, the temperature changed to $$T_2$$. The pressure of gas is found to be 10 $$atm$$. Find out the value of $$T_2$$:
    Solution
    $$6.023 \times 10^{23}$$ molecules corresponds to 1 $$mole$$
    $$2.41 \times 10^{23}$$ molecules corresponds to $$\displaystyle =\frac{2.41 \times 10^{23}}{6.023 \times 10^{23}}=0.4\ moles$$
    Case (i)
    $$P_1 V_1 = n_1 R_1 T_1$$
    $$8(2.5)=n_1 (0.08) (300)$$
    $$n_1 = \displaystyle \frac{8(2.5)}{(0.08)(300)}= 0.833\ moles$$
    Total moles $$n_2 = 0.833 +0.4$$
    $$n_2 = 1.233$$
    Case (ii)
    $$P_2V_2=n_2R  T_2$$
    $$10(2.5)=1.233 (0.08) T_2$$
    $$T_2=253.4$$
  • Question 5
    1 / -0
    In the kinetic theory of gases, it is assumed that molecular collisions are :
    Solution
    The assumption of the kinetic molecular theory
    1. A gas consists of molecules in constant random motion.
    2. Gas molecules influence each other only by collision, they exert no other forces on each other.
    3.all collisions between gas molecules are perfectly elastic and short in duration. Thus all KE is conserved
    4. The volume actually occupied by the molecules by the gas is negligibly small.
  • Question 6
    1 / -0
    According to kinetic theory of gases, for a diatomic gas
    Solution
    According to kinetic theory, mean translational kinetic energy $$(\epsilon)=\dfrac {3}{2}k_BT$$.

    Where, $$k_B$$ is the Boltzmann's constant. 

    Thus, kinetic energy is directly proportional to temperature.

    Hence, the correct option is $$C$$
  • Question 7
    1 / -0
    When an ideal gas undergoes unrestrained expansion, no cooling occurs because of the molecules:
    Solution
    According to postulates of kinetic theory, there is no intermolecular attractions or repulsions between the molecules of ideal gases.
    Thus when an ideal gas undergoes unrestrained expansion, no cooling occurs because the molecules do not exert any attractive forces on each other.
  • Question 8
    1 / -0
    The density of neon will be highest at:
    Solution
    The ideal gas equation is $$PV=nRT=\cfrac {w}{M}RT$$.

    The above equation is rearranged to obtain 

    $$\Rightarrow PM=\left(\cfrac {w}{V}\right)RT=dRT $$. 

    Here, d represents the density.

    Hence, the expression for the density becomes 

    $$\Rightarrow d=\cfrac {PM}{RT}$$.

    Thus low temperature and high pressure result in greater density.

    Thus, the density of neon will be highest at $$0^oC, 2  \:atm$$.

    Hence, the correct option is $$\text{B}$$
  • Question 9
    1 / -0
    Which of the following changes cannot increase the volume of a gas by 4 times?
    Solution
    According to gas equation,

    $$\cfrac {P_1V_1}{T_1}=\cfrac {P_2V_2}{T_2}$$ and $$V_2=\cfrac {P_1V_1T_2}{T_1P_2}$$

    In Option (A),
    $$\therefore T_2=2T_1, P_2=\cfrac {1}{2}P_1$$

    $$V_2=\cfrac {P_1}{P_2}\times \cfrac {T_2}{T_1}\times V_1$$

    $$V_2=2\times 2\times V_1=4V_1$$

    In option (B),

    $$P_2=P_1, T_2=4T_1$$

    $$V_1=\cfrac {P_1}{P_2}\times \cfrac {T_2}{T_1}\times V_1$$

    $$=1\times 4V_1=4V_1$$

    In option (C),

    $$T_2=\cfrac {1}{2}T_1, P_2=2P_1$$

    $$V_2=\cfrac {P_1}{P_2}\times \cfrac {T_2}{T_1}\times V_1$$

    $$=\cfrac {1}{2}\times \cfrac {1}{2}V_1=\cfrac {V}{4}$$

    In option (D),

    $$T_2=T_1, P_2=\cfrac {1}{4}P_1$$

    $$V_2=\cfrac {P_1}{P_2}\times \cfrac {T_2}{T_1}\times V_1$$

    $$=4\times 1\times V_1=4V_1$$

    In option C volume doesn't become 4 times.
    Option C is correct.
  • Question 10
    1 / -0
    The volume occupied by 11 grams of $$CO_2$$ gas at STP is:
    Solution
    As we know,
    One mole of a gaseous substance occupies $$22.4\ litre$$.

    Mole of $$CO_2 = 11/44 = 1/4$$ mole

    Volume occupied $$=\dfrac{22.4}{4} = 5.6\ litre$$
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