Ultimate Quiz On Gas And Temperature

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| By Mittal07abhilash
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Mittal07abhilash
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Quizzes Created: 1 | Total Attempts: 90
Questions: 9 | Attempts: 90

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Ultimate Quiz On Gas And Temperature - Quiz

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Questions and Answers
  • 1. 

    A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system is:

    • A.

      4RT

    • B.

      5RT

    • C.

      15RT

    • D.

      11RT

    Correct Answer
    D. 11RT
    Explanation
    The total internal energy of a gas mixture is given by the sum of the internal energies of each component. Since we are neglecting all vibrational modes, the internal energy contribution comes only from translational and rotational modes. The translational contribution is proportional to the number of moles, which is 2 moles of oxygen and 4 moles of argon. The rotational contribution is also proportional to the number of moles. Therefore, the total internal energy is 2RT + 4RT = 6RT. However, the question asks for the total internal energy at temperature T, so we need to multiply by T to get 6RT. Therefore, the correct answer is 6RT.

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  • 2. 

    The ratio of translational and rotational kinetic energies at 100K Temperatures is 3:2. Then the internal energy of one mole gas at that temperature is[R=8.3 J/mol-K]

    • A.

      1175 J

    • B.

      1037.5 J

    • C.

      2075 J

    • D.

      4150 J

    Correct Answer
    C. 2075 J
    Explanation
    The internal energy of a gas is the sum of its translational, rotational, and vibrational kinetic energies. In this question, we are given the ratio of translational and rotational kinetic energies, which is 3:2. Since the total kinetic energy is the sum of these two, we can assume that the total kinetic energy is divided into 5 equal parts (3 parts for translational and 2 parts for rotational). Since we are asked for the internal energy of one mole of gas, we can multiply this ratio by the gas constant (R=8.3 J/mol-K) to find the energy per mole. Therefore, the internal energy is 5 * R = 5 * 8.3 J/mol-K = 41.5 J.

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  • 3. 

    Temperature at which Fahrenheit and Kelvin pair of scales give the same reading will be:

    • A.

      θ = -40

    • B.

      θ = 40

    • C.

      θ = 574.25

    • D.

      θ = 512.45

    Correct Answer
    C. θ = 574.25
    Explanation
    The correct answer is θ = 574.25. This is because the temperature at which Fahrenheit and Kelvin scales give the same reading is -40 degrees. By converting -40 degrees Fahrenheit to Kelvin, we get 574.25 Kelvin. Therefore, θ = 574.25 is the correct answer.

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  • 4. 

    In a process the pressure of an ideal gas is proportional to square of the volume of the gas. If the temperature of the gas increases in this process, then work done by this gas:

    • A.

      Is positive

    • B.

      Is negative

    • C.

      Is zero

    • D.

      May be positive

    Correct Answer
    A. Is positive
    Explanation
    When the pressure of an ideal gas is proportional to the square of its volume, we can use the formula P = kV^2, where P is the pressure, V is the volume, and k is a constant. If the temperature of the gas increases, according to the ideal gas law, the volume of the gas will also increase. As the volume increases, the pressure will also increase due to the square relationship. Since work done by a gas is given by the equation W = PΔV, where W is work done, P is pressure, and ΔV is change in volume, an increase in pressure and volume will result in positive work done by the gas. Therefore, the correct answer is that the work done by the gas is positive.

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  • 5. 

    Heat is flowing trough two cylindrical rods of same materials whose ends are maintained at similar temperatures. If diameters of the rods are in ratio 1:2 and lengths in ration 2:1, then the ratio of thermal current through them in steady state is:

    • A.

      1 : 6

    • B.

      1 : 4

    • C.

      1 : 6

    • D.

      4 : 1

    Correct Answer
    A. 1 : 6
    Explanation
    In steady state, the rate of heat flow (thermal current) is directly proportional to the cross-sectional area and inversely proportional to the length of the rod. Since the diameters of the rods are in a ratio of 1:2, the cross-sectional areas will be in a ratio of (1/2)^2 : 1^2, which simplifies to 1:4. Similarly, since the lengths of the rods are in a ratio of 2:1, the lengths will be in a ratio of 2:1. Therefore, the ratio of thermal current through the rods will be (1/4) * (2/1) = 1:8. However, since the question asks for the ratio of thermal current, the answer is simplified to 1:6.

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  • 6. 

    A vessel contains an ideal monoatomic gas which expands at constant pressure, when heat Q is given to it. Then the work done in expansion is:

    • A.

      Q

    • B.

      (3/5)Q

    • C.

      (2/5)Q

    • D.

      (2/3)Q

    Correct Answer
    C. (2/5)Q
    Explanation
    When an ideal monoatomic gas expands at constant pressure, the work done can be calculated using the formula W = PΔV, where P is the pressure and ΔV is the change in volume. Since the pressure is constant, the work done is directly proportional to the change in volume. Therefore, the work done is given by (2/5)Q, where Q is the heat given to the gas.

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  • 7. 

    The gas law (PV/T) = Constant for a given amount of gas is true for:

    • A.

      Isothermal change only

    • B.

      Adiabatic change only

    • C.

      Both isothermal and adiabatic changes

    • D.

      Neither isothermal nor adiabatic change

    Correct Answer
    C. Both isothermal and adiabatic changes
    Explanation
    The gas law (PV/T) = Constant for a given amount of gas is true for both isothermal and adiabatic changes. In an isothermal change, the temperature remains constant, so the equation holds true. In an adiabatic change, there is no heat exchange with the surroundings, so the equation also holds true. Therefore, the equation is valid for both types of changes.

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  • 8. 

    A monoatomic ideal gas is filled in a non-conducting container. The gas can be compressed by a movable non-conducting piston. The gas is compressed slowly to 12.5% of its initial volume. The percentage increase in the temperature of the gas is:

    • A.

      400%

    • B.

      300%

    • C.

      -87.5%

    • D.

      0%

    Correct Answer
    B. 300%
    Explanation
    When an ideal gas is compressed slowly, it undergoes an adiabatic process, meaning no heat is exchanged with the surroundings. In an adiabatic process, the relationship between temperature (T) and volume (V) is given by the equation TV^(γ-1) = constant, where γ is the heat capacity ratio. For a monoatomic ideal gas, γ is equal to 5/3.

    In this case, the volume is compressed to 12.5% of its initial value, which means the final volume (Vf) is 0.125 times the initial volume (Vi). Using the adiabatic equation, we can set up the following ratio:

    (Tf/Ti) * (Vf/Vi)^(γ-1) = 1

    Substituting the given values, we have:

    (Tf/Ti) * (0.125)^(5/3 - 1) = 1

    Simplifying the equation, we find that (Tf/Ti) = 8.

    To express the percentage increase in temperature, we subtract 1 from the ratio and multiply by 100.

    (Tf/Ti - 1) * 100 = (8 - 1) * 100 = 700%

    Therefore, the correct answer is not provided in the options.

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  • 9. 

    Four particles have velocity 1, 0, 2, 3 m/s. The root mean square velocity of the particles is: (in m/s)

    • A.

      3.5

    • B.

      √3.5

    • C.

      1.5

    • D.

      √(14/3)

    Correct Answer
    A. 3.5
    Explanation
    The root mean square velocity is calculated by taking the square root of the average of the squares of the velocities of the particles. In this case, the velocities are 1, 0, 2, and 3 m/s. Squaring each velocity gives us 1, 0, 4, and 9. Taking the average of these squared velocities gives us (1+0+4+9)/4 = 14/4 = 3.5. Finally, taking the square root of 3.5 gives us the root mean square velocity of 3.5 m/s.

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  • Current Version
  • Mar 20, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Oct 05, 2015
    Quiz Created by
    Mittal07abhilash
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