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Kinetic Theory Test - 90

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Kinetic Theory Test - 90
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  • Question 1
    1 / -0
    The molar heat capacity in  a process of a  diatomic gas if does a work of $$\frac{Q}{4}$$  when a heat of  Q is supplied to is 
    Solution

  • Question 2
    1 / -0
    One mole of an ideal gas passes through a process where pressure and volume obey the relation $$P = P_0 \left[1 - \dfrac{1}{2} \left(\dfrac{V_0}{V} \right)^2 \right]$$. Here $$P_0$$ and $$V_0$$ are constants . Calculate the change in the temperature of the gas if its volume change from $$V_o $$ to $$2 V_o$$.
    Solution

  • Question 3
    1 / -0
    Three lawn chairs, one made up of aluminium (heat capacity $$=0.90J/K-g$$), one of iron (heat capacity $$=0.45J/K-g$$) and one of tin (heat capacity $$=0.60J/K-g$$) are painted of the same colour. On a sunny day which chair will be hotter to sie?
    Solution

  • Question 4
    1 / -0

    Directions For Questions

    An intimate mixture of hydrogen gas and the theoretical amount of air at $$25^{o}C$$ and a total pressure of $$1\ atm$$, is exposed in a closed rigid vessel. If the process occurs under adiabatic condition, then using the following data, answer the following question :
    Given ; $$ (i) C_{P,m} = 8.3\ cal/K-mol\ (ii)\ C_{P, m}= 11.3\ cal/K-mol\ (iii)\ \triangle_{f} H [H_{2}O (g)]= -57.8\ kcal $$$$ (iv)\ Air \ contain \ 20\% O_{2}$$  and $$80\% \ N_{2}$$, by volume 

    ...view full instructions

    The values of $$C_{P, m}$$ of $$N_{2}(g)$$ and $$H_{2}O (g)$$ (in $$cal/K-mol$$) should be 
    Solution

  • Question 5
    1 / -0
    An average human produces about $$10MJ$$ of heat each day through metabolic activity. If a human body were an isolated system of mass $$80kg$$ with the heat capacity of water, what temperature rise would be body experience? Heat capacity of water $$=4.2J/K-g$$
    Solution
    heat produced $$=10MJ=10\times {10}^{6}J$$

    mass $$=80kg$$

    $$C=4.2J/K-g$$

    we have relation

    $$Q=mc\Delta T$$

    $$10\times {10}^{6}=80\times {10}^{3}\times 4.2\times \Delta T$$

    $$\Delta T=\cfrac{10\times {10}^{6}}{80\times {10}^{3}\times 4.2}$$

    $$\Delta T=0.02976\times {10}^{3}$$

    $$\Delta T={29.76} K $$

    Option A is correct.
  • Question 6
    1 / -0
    The heat capacity of liquid water is $$75.6J/K-mol$$, while the enthalpy of fusion of ice is $$6.0kJ/mol$$. What is the smallest number of ice cubes at $${0}^{o}C$$, each containing $$9.0g$$ of water, needed to cool $$500g$$ of liquid water from $${20}^{o}C$$ to $${0}^{o}C$$?
    Solution
    $$500g$$ of $${H}_{2}O$$

    moles of $${H}_{2}O=\cfrac{500g}{18g/mol}=27.78moles$$

    $$Q$$ heat generated $$=mc\Delta T$$

    $$=27.78 mol \times 75.6 J/K- mol\times (293-273)K$$

    $$=42,000J$$

    $$Q=42,000\times \dfrac{1kJ}{1000J}=42kJ/mol$$

    Heat of fusion $$=6kJ/mol$$

    But we have $$9g$$ $${H}_{2}O$$

    moles $$=\cfrac{9g}{18g/mol}=0.5mol$$

    So, $$0.5$$ mole of ice (1 ice cube) will have $$3\ kJ$$ of energy.

    500g of liquid water contains 42kJ of energy.

    9g of water at $$0^oC$$ = 1 ice cube contains 3kJ energy.

    So, the number of ice cubes required $$=\dfrac{Total\ energy\ of\ liquid\ water}{energy\ per\ ice\ cube}=\cfrac{42}{3}=14$$

    Hence, the correct option is $$C$$
  • Question 7
    1 / -0
    One gram-mole of an ideal gas $$A$$ with the ratio of constant pressure and constant volume specific heats, $$\gamma _A = \dfrac{5}{3}$$ is mixed with $$n$$ gram-moles of another ideal gas $$B$$ with $$\gamma_B = \dfrac{7}{5}$$. If the $$\gamma$$ for the mixture is $$\dfrac{19}{13}$$ what will be the value of $$n$$?
    Solution

  • Question 8
    1 / -0
    In a model of chlorine $$(Cl_2)$$, two $$Cl$$ atoms are rotated about their centre of mass as shown. Here the two $$'CI'$$ atoms are $$2\times 10^{-10}m$$ apart and angular speed $$\omega =2\times 10^{12}\ rad/s$$. If the molar mass of chlorine is $$70\ g/mol$$, then what is the rotational kinetic energy of one $$Cl_2$$ molecule?

    Solution

  • Question 9
    1 / -0
    If at NTP, velocity of sound in a gas is $$1150\ m/s$$, then the $$rms$$ velocity of gas molecules at $$NTP$$ is :( Given : $$R=8.3\ joule/mol/K, C_p=4.8\ cal/mol/K$$)
    Solution

  • Question 10
    1 / -0
    We would like to increase the length of a $$15\ cm$$ long copper rod of cross-section $$4\ mm^2$$ by $$1\ mm$$. The energy absorbed by the rod if it is heated is $$E_1$$. The energy absorbed by the rod if it is stretched slowly is $$E_2$$. Then $$E_1 /E_2$$ is:
    Solution

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