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Gaseous States Test - 13

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Gaseous States Test - 13
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Weekly Quiz Competition
  • Question 1
    1 / -0

    Pressure of a mixture of 4g of O2 ​ and 2g of H2 ​ confined in a bulb of 1 litre at 0oC is:

    Solution

     

  • Question 2
    1 / -0

    A sample of gas occupies 100 mL at 27C and 740 mm pressure. When its volume is changed to 80 mL at 740 mm pressure, the temperature of the gas will be :

    Solution

     

  • Question 3
    1 / -0

    A sample of a given mass of gas at a constant temperature occupies 95cm3 under a pressure of 9.962×104Nm2. At the same temperature its volume at a pressure of 10.13×104Nm−2 is: 

    Solution

    According to boyle's law At constant temperature

     

  • Question 4
    1 / -0

    The density of a gas A is three times of a gas B,if the molecular mass of A is M, the molecular mass of B is

    Solution

     

  • Question 5
    1 / -0

    Pure hydrogen sulphide is stored in a tank of 100 litre capacity at 20C and 2 atm pressure. The mass of the gas will be :

    [S = 32,H = 1].

    Solution

    Here using ideal gas equation,

    given :

    V = 100L

    P = 2atm

    M(molar mass) = 34gmol−1

    T = 293K

    R = 0.0821L atm mol−1T−1

    Solution: P×V=n×R×T

    Substituting the given values,

    n = 8.31mol

    But, n = M/W

    W = 8.31 × 34

    W = 282.54g

     

  • Question 6
    1 / -0

    For an ideal gas, a number of moles per liter in terms of its pressure P, gas constant R and temperature T is:

    Solution

    According to Ideal Gas Equation, PV = nRT

    P is the pressure, V is the volume, n is the number of moles, R is gas constant and T is the temperature. 

    Therefore number of moles per litre, 

    n​/v =  P/RT

     

  • Question 7
    1 / -0

    If 2 moles of an ideal gas at 546K occupy volume 44.8L, then pressure must be :

    Solution

    From gas equation, PV = nRT

    Given that :

    V = 44.8L
    n = 2
    R = 0.0821 L atm/mol/K
    T = 546K

     

  • Question 8
    1 / -0

    Select one correct statement. In the gas equation, PV = nRT

    Solution

    PV = nRT

    where P= pressure of the gas;  

    V=volume of the gas;  

    n= Number of Moles;  

    T=Absolute temperature; 

    R=Ideal Gas constant also known as Boltzmann Constant = 0.082057 L atm K−1 mol−1.

    Here V ∝ n 

    Hence, V is the volume of n moles of gas

     

  • Question 9
    1 / -0

    A gas can be liquefied ______.

    Solution

    A gas can be liquefied below its critical temperature.

    For example, the critical temperature Tc​ for CO2​, O2​ and H2​ are 31.1 0C, −118.8 0C and −240 0C respectively. 

    Thus, CO2​ gas cannot be liquefied above 31.10C however high the pressure may be applied.

     

  • Question 10
    1 / -0

    At 27C, the ratio of rms velocities of ozone to oxygen is

    Solution

     

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