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Kinetic Theory Test - 34

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Kinetic Theory Test - 34
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  • Question 1
    1 / -0
    The ratio of kinetic energy to potential energy for solids is
    Solution
    Since the particles are tightly packed in solids, so they possess more potential energy than kinetic energy, which is released on heating the solid.
  • Question 2
    1 / -0
    The correct curve for a stable diatomic molecule is
    Solution
    As shown in diagram , at a particular distance between atoms(equilibrium state), energy will be minimum which means that gas molecule is at stable configuration.As distance increases or decreases,electrostatic  force between atoms increases which make molecule unstable and energy also increases.

  • Question 3
    1 / -0
    A vessel of volume 4 litres contains a mixture of 8 g of O$$_2$$, 14 g of N$$_2$$ and 22g of CO$$_2$$ at 27$$^o$$C. The pressure exerted by the mixture is
    Solution
    Moles of $$O_2=\dfrac{8}{32}=\dfrac{1}{4}$$
    Moles of $$N_2=\dfrac{14}{28}=\dfrac{1}{2}$$
    Moles of $$CO_2=\dfrac{22}{44}=\dfrac{1}{2}$$
    Thus total moles in the mixture=$$\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{5}{4}$$
    Using Ideal Gas Equation,
    $$PV=nRT$$
    $$\implies P=\dfrac{nRT}{V}=\dfrac{\dfrac{5}{4}\times 8.3\times 300}{4\times 10^{-3}}N/m^2\approx 7.79\times 10^{5}N/m^2$$
  • Question 4
    1 / -0
    A vessel contains $$1$$ mole of O$$_2$$ (molar mass $$ 32 gm$$) at a temperature $$T$$. The pressure is $$P$$. An identical vessel containing $$1$$ mole of He (molar mass $$4 gm$$) at a temperature $$2T$$ has a pressure :
    Solution
    Using $$PV=nRT$$
    $$\therefore P_{He} = 2 P$$ as temperature of He is doubled.
  • Question 5
    1 / -0
    The temperature of the sun, if pressure is $$1.4 \times 10^9$$ atm, density is $$1.4 gcm^{-3}$$ and average molecular weight is 2, will be
    $$[Given\  R = 8.4  J  mol^{-1} K^{-1}]$$
    Solution
    $$PV = nRT     n = \displaystyle \frac{m}{M} $$ and $$\rho = \displaystyle \frac{m}{V}$$
    or
    $$\displaystyle T = \frac{PV}{nR} = \frac{PM}{\rho R}$$
    $$= \displaystyle \frac{1.4 \times 10^9 \times 1.01 \times 10^5 \times 2 \times 10^{-3}}{1.4 \times 10^3 \times 8.4}$$
    $$= 2.4 \times 10^7  K$$
  • Question 6
    1 / -0

    Directions For Questions

    An automobile tyre has volume $$0.015 m^3$$ on a cold day when temperature is $$5^o C$$ and atmospheric pressure is $$1.02\ atm$$. Under these conditions the gauge pressure is $$1.7\ atm\ (25\ lb/in^2)$$. After the car is driven on a highway for 30 minutes the temperature of the air in the tyres rises to $$45^oC$$ and volume to $$0.0159 m^3$$.

    ...view full instructions

    What is new gauge pressure?
    Solution
    $$\displaystyle \frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$$
    or $$\displaystyle P_2 = \frac{2.72 \times 150 \times 318}{278 \times 159}$$
    $$= 2.94 atm$$
    Gauge pressure $$= 2.94 - 1.02$$
    $$= 1.92 atm$$
  • Question 7
    1 / -0
    A barometer tube 90 cm long contains some air above mercury. The reading is 74.8 cm when true atmospheric pressure is 76 cm and temperature is 30$$^o$$ C. If the reading is observed to be 75.4 cm on a day when temperature is 10$$^o$$C, then find the true pressure.
    Solution
    Let A be the area of the cross-section
    $$V_1 = (90 - 74.8) A = 15. 2 A  cm^3$$
    $$P_1 = 76 - 74.8 = 1.2$$ cm of Hg
    $$P_2 = (P - 75.4)$$ cm of Hg
    $$V_2 = (90 - 75.4) A = 14.6 A  cm^3$$
    $$\displaystyle \frac{P_1V_1}{T_1} = \frac{P_2 V_2}{T_2} = \frac{1.2 \times 15.2 \times 283}{303 \times 14.6} = 75.4 + 1.17$$
    $$= 76.57$$ cm of Hg
  • Question 8
    1 / -0
    Equal masses of N$$_2$$ and O$$_2$$ gases are filled in vessels A and B. The volume of vessel B is double of A. The ratio of pressure in vessel A and B will be
    Solution
    Moles of $$N_2=n_A\dfrac{m}{28}$$

    Moles of $$O_2=n_B\dfrac{m}{32}$$
    Given that $$V_B=2V_A$$
    $$P\propto \dfrac{n}{V}$$
    $$\implies \dfrac{P_A}{P_B}=\dfrac{n_A}{n_B}\dfrac{V_B}{V_A}=\dfrac{16}{7}$$
  • Question 9
    1 / -0
    At what temperature the average translational KE of the molecules of a gas will become equal to the KE of an electron accelerated from rest through 1 V potential difference?
    Solution
    $$\displaystyle \frac{3}{2} KT = 1 eV = 1.6 \times 10^{-19} J$$
    or
    $$\displaystyle T = \frac{2 \times 1.6 \times 10^{-19}}{3 \times 1.38 \times 10^{-23}} = 7730 K=7.73\times10^3K$$
  • Question 10
    1 / -0
    One mole of an ideal gas undergoes a process $$P = \displaystyle \frac{P_o}{1 + \left ( \displaystyle\frac{V}{V_o} \right )^2}$$, where $$P_o$$ and $$V_o$$ are constants. Find the temperature of the gas when $$V = V_o$$.
    Solution
    Since the gas is ideal, therefore $$P_o V_o = RT$$

    Using the relation given when $$V = V_o$$ 

    $$P = \displaystyle \frac{P_o}{2}$$

    Thus we get

    $$ \displaystyle \frac{P_o}{2} (V_o) = RT$$

    $$T = \displaystyle \frac{P_oV_o}{2R}$$.
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