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Kinetic Theory Test - 68

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Kinetic Theory Test - 68
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  • Question 1
    1 / -0
    $$2$$ mole ideal He gas and $$3$$ mole ideal $$H_2$$ gas at constant volume find out $$C_v$$ of mixture 
    Solution
    The given mixture of two moles of monoatomic gas mixed with three mole of diatomic gas can be given as 
    $$C=\dfrac{n_1C_{v_1}+n_2C_{v_2}}{n_1+n_2}$$

        $$=\dfrac{2\times \dfrac{3}{2}R+3\times \dfrac{5}{2}R}{2+3}$$

        $$=\dfrac{3R+15\dfrac{R}{2}}{5}=\left(\dfrac{21 R}{10}\right)$$
  • Question 2
    1 / -0
    In a jar having a mixture of $$H_{2}$$ and He 
  • Question 3
    1 / -0
    In Maxwell's speed distribution curve, for N2N2K will be?
    Solution
    Magnitude of relative velocity, 
    $$|V_{rev}|=\sqrt{V^2+V^2-2(V)(V)\cos\theta}=2V\left|\sin\dfrac{\theta}{2}\right|$$

    Average of Relative velocity 
    $$(V_{rev})=\dfrac{\displaystyle\int^{\pi}_02V\left|\sin\dfrac{\theta}{2}\right|d\theta}{\displaystyle\int^{\pi}_0d\theta}=\dfrac{4V}{\pi}$$
    $$(V_{rev})=\dfrac{4}{\pi}V_{average}=\dfrac{4}{\pi}\sqrt{\dfrac{8RT}{\pi m_0}}=\dfrac{4}{\pi}\sqrt{\dfrac{8\times 8.3\times 300}{3.14\times 28\times 10^{-3}}}=606$$m/sec.

  • Question 4
    1 / -0
    r.m.s. speed of ideal gas at $$127^oC$$ is $$200m/s$$, the r.m.s. speed of same ideal gas at temperature $$227^oC$$ is:
    Solution
    $$v = \sqrt{\dfrac{3RT}{M}}$$
    $$\dfrac{v_1}{v_2} = \sqrt{\dfrac{T_1}{T_2}} = \dfrac{200}{v_2} = \sqrt{\dfrac{400}{500}}$$
    $$v_2 = 100\sqrt{5}$$
  • Question 5
    1 / -0
    At constant pressure, the heat of formation of a compound is not dependent on temperature, when: 
    Solution
    At constant pressure, the enthalpy of element or compound changes with temperature according to the formula:

    $$\Delta H=\Delta C_p T$$

    So, if $$\Delta C_p=0$$ then $$\Delta H=0$$

    Thus, the heat of formation of the compound becomes independent of temperature.
  • Question 6
    1 / -0
    Pressure of an ideal gas is increased by keeping temperature constant. The kinetic energy of molecules.
    Solution
    Since the pressure is being increased, keeping the temperature constant, this basically implies that you are squeezing the volume of the container in which the gas is kept (a direct consequence of Boyle’s Law). Since, the volume of the container has decreased, the dimensions over which the molecules of the gas can move before colliding with the walls has also decreased. Hence, there will be greater number of collisions per unit time. But that is all that would happen.
    Since the temperature is essentially held constant, and it is a common knowledge that the root mean square speed of the molecules of an ideal gas is purely a function of temperature alone, as given by V = (3RT/M)0.5, so the kinetic energy of the molecules will remain constant. There may be greater amount of energy exchange and transfer between the molecules, due to reduction in mean free path. But, the overall kinetic energy of the gas as a whole, remains the same.

  • Question 7
    1 / -0
    Which row correctly describes the ordering and motion of the molecules in liquid water and in ice when both are at a temperature of $${0}^{o}C$$
    Solution
    Since the average speed of molecule depends only on temperature . So, both have same speed.
    Now, the ice is in solid motion , so the molecules of ice are arranged in a regular pattern. 
    Therefore correct option is A
  • Question 8
    1 / -0
    A volume of $$2.5\ L$$ of a sample of a gas at $$27^o$$C and $$1$$ bar pressure is compressed to a volume of $$500\ ml$$ keeping the temperature constant, the percentage increase in pressure is?
    Solution
    $$V_1=2.5\ L $$

    $$T_1=27^oC=300K$$

    $$P_1=1$$ bar

    $$V_2=500$$ml$$=0.5\ L$$

    $$T_2=300\ K$$

    $$P_2=?$$

    $$P_1V_1=P_2V_2$$

    $$2.5\times 1=P_2\times 0.5$$

    $$\therefore P_2=5$$ bar


    Now $$\%$$ increase in pressure$$=\dfrac{P_2-P_1}{P_1}$$
    $$=\dfrac{5-1}{1}\times 100$$
    $$=400\%$$.
  • Question 9
    1 / -0
    If the temperature of $$3$$ moles of helium gas is increased by $$2\ K$$, then the change in the internal energy of helium gas is :
    Solution

  • Question 10
    1 / -0
    In the case of solid, number of degrees of freedom is :
    Solution

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