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Thermal Properties of Matter Test - 23

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Thermal Properties of Matter Test - 23
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  • Question 1
    1 / -0

    The temperature at which the volume of ideal gas is zero is:

    Solution
    From ideal gas equation, PV=nRTPV = nRT
    This gives V=nRTPV = \dfrac{nRT}{P}
    Now V tends to zero when T tends to zero where T is in Kelvin
    So, volume becomes 0 only when temperature = 0K=273 0 K = -273 ^{\circ}C
  • Question 2
    1 / -0

    Two identical tumblers are filled fully with water at a certain temperature. One of the tumblers is warmed up and the other cooled down. But it is observed that the water overflows from both the tumblers. The temperature at which they are fully filled is

    Solution
    Water has anomalous expansion from 404^0C(2770277^0K) to 000^0C(2730273^0K).
    Water expands when heated from 404^0C(2770277^0K) onwards.
    Since water overflows both the tumblers, the temp at which they are fully filled is 2770277^0K
  • Question 3
    1 / -0

    At constant pressure, density of a gas is :

    Solution

    Hint: Density of a substance is given by ρ =mV\rho  = \dfrac{m}{V} , where mmis the mass of the substance and VV is its volume.

    Correction Option: B

    Explanation for correct option:

    • By ideal gas equation: PV=nRTPV = nRTwhere PP,VVTT and nn  are the pressure, volume, temperature and moles of the gas and RR is the universal gas constant.

    PVα nRT \Rightarrow PV \alpha  nRT

    PV=mMRT \Rightarrow PV = \dfrac{m}{M} RTwhere mm and MM are weight and molecular weight of the gas.

    P=mVRTM \Rightarrow P = \dfrac{m}{V}\dfrac{{RT}}{M}

    Here, ρ =mV\rho  = \dfrac{m}{V} and R=RMR' = \dfrac{R}{M} where RR' is the specific gas constant for the gas.

    P=ρRT \Rightarrow P = \rho R'T

    ρ =PRT \Rightarrow \rho  = \dfrac{P}{{R'T}}

    ρ α   1T\therefore \rho  \alpha     \dfrac{1}{T}

    Hence, at constant pressure, density of a gas is inversely proportional to absolute temperature.

  • Question 4
    1 / -0

    If the slope of P-T graph for a given mass of a gas increases, then the volume of the gas

    Solution
    From ideal gas equation, PV = nRT = wMRT\frac{w}{M}RT
    slope of P-T curve = wRMV\frac{wR}{MV}
    since V = constant and w = constant, we get that the slope is inversely proportional to molecular weight
    Steeper the line, more is slope, less is volume.
    So as slope increases, volume decreases
  • Question 5
    1 / -0

    An ideal gas is that which

    Solution
    The concept of liquefaction of gases works on the principle of intermolecular forces of attraction between the gas molecules. Gases liquefy when these forces increase between the molecules, binding them together. But an ideal gas is one where the molecules do not influence each other, the explanation comes from Joule Thompson effect.
  • Question 6
    1 / -0

    The temperature at which a body does not radiate energy

    Solution
    Energy radiated becomes 00 only when absolute temperature becomes 00.
  • Question 7
    1 / -0

    A box contains x molecules of a gas. How will the pressure of the gas be affected if the number of molecules is made 2x?

    Solution
    we know that from ideal gas equation, PV = nRT
    now, n = number of moles of gas = NNA\dfrac{N}{N_{A}}, where N = number of molecules of the gas and NAN_{A} is the Avogadro's number
    So, putting the value of n in ideal gas equation, we get
    PV=NNARTPV = \dfrac{N}{N_{A}}RT
    this means that pressure of the gas is proportional to the number of molecules of the gas at constant temperature and volume
    So, if we double the number of molecules of the gas (2x), then pressure will also double
    So, C is the correct answer.
  • Question 8
    1 / -0

    In a gas equation, PV = RT, V refers to the volume of:

    Solution
    Given PV = RT
    Comparing with the ideal gas equation, PV = nRT, we
    find that here n = 1
    So, number of moles = 1
    Hence we must refer to molar volume
    So, C is the correct answer.
  • Question 9
    1 / -0

    The relation between volume V, pressure P and absolute temperature T of an ideal gas is PV = xT, where x is a constant. The value of x depend upon

    Solution
    From ideal gas equation we know that PV = nRT
    We are given that PV = xT
    Comparing the 2, we get that x = nR
    Now, n = number of moles of the gas = NNA\frac{N}{N_{A}}, where N is the number of molecules of the gas and NAN_{A} is the avogadro's number
    So, x = NNAR\frac{N}{N_{A}}R
    Clearly NAN_{A} and R are universal constants
    So, x depends on N = number of molecules of the gas.
    So, D is the correct answer.
  • Question 10
    1 / -0

    The universal gas constant has the units

    Solution
    from ideal gas equation, PV = nRT
    now, the dimensions of PV is same as dimension of energy (or dimension of work)
    So, dimension of PV is ergs
    and the dimension of n is mol, dimension of T is Kelvin
    So, dimension of R = $$\dfrac{dimension\  of\  PV}{dimension\  of\  n  \times  dimension\ of\ Temperature}$$
    =ergmolK = \dfrac{erg}{mol-K}
    so, B is the correct answer.
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