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Thermodynamics Test - 25

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Thermodynamics Test - 25
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  • Question 1
    1 / -0
    Properties of substances like pressure, temperature and density, in thermodynamic coordinates are 
    Solution
    Point Function: They depend on the state only, and not on how a system reaches that state. They are also known as state functions. All properties of a substance are point functions. For example- pressure, temperature and density. 
    In contrast,
    Path function: Their magnitudes depend on the path followed during a process as well as the end states. For example- work done in moving from one point to the other.
  • Question 2
    1 / -0
    Which of the following property is not a thermodynamic property of the system?
    Solution
    A thermodynamic property is that, which is measurable and whose value describes the state of system. Out of the given quantities, heat doesn't describe the state of a system so it is not a thermodynamic property because a system doesn't contain heat but only can transfer heat.
  • Question 3
    1 / -0
    Which of the following is the property of a system 
    Solution
    All the given quantities in the question, describe the state of system, therefore they are the properties of a system.
  • Question 4
    1 / -0
    Mixture of ice and water is form a 
    Solution
    Mixture of ice and water is a heterogeneous system because their chemical compositions are same but the physical properties are completely different. 
  • Question 5
    1 / -0
    Molar heat capacity of a gas does not depend on.
    Solution
    The heat capacity specifies the heat needed to raise a certain amount of a substance by 1 K. For a gas, the molar heat capacity C is the heat required to increase the temperature of 1 mole of gas by 1 K. Important: The heat capacity depends on whether the heat is added at constant volume or constant pressure.
    Hence the Molar heat capacity does not depends on its temperature or say gas temperture 
  • Question 6
    1 / -0
    Fill in the blank. 
    Thermodynamics is the branch of science concerned with ____ and ________ and their relation to energy and work. 
    Solution
    Thermodynamics is the branch of physics concerned with heat and temperature and their relation to energy and work. In thermodynamics, we study about the concepts of heat, temperature and relation between heat energy and mechanical energy.
  • Question 7
    1 / -0
    A cycle tyre bursts suddenly. What is the type of this process?
    Solution
    Any process  is adiabatic is rapid such that there should not be any heat transfer between the system and it's surroundings. When a tyre bursts suddenly, the expansion takes place instantly. This leads to decrease in temperature inside. As such, the higher temperature air outside will transfer heat to it. This heat transfer is not rapid and doesn't take place instantly, unlike the expansion, which is instantaneous. Heat transfer takes place after the bursting, due to which one can  consider that there is almost no energy  exchange during the actual process. Thus process is adiabatic.
  • Question 8
    1 / -0
    Which of the following system exchanges only energy but not matter?
    Solution
    When there is no exchange of matter between surrounding and a thermodynamic system but system allows the exchange of energy as heat or work, then the system is called a closed system.
  • Question 9
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    According to given figure which of the following options can be true ?

    Solution
    The second law of thermodynamics states , that it is impossible to construct a heat engine , operating in cycle , which extracts heat and can convert it all to useful work .In other words , it is impossible for a heat engine to convert heat completely in work .
    The efficiency of a heat engine is the ratio of net work done per cycle by engine to the heat absorbed per cycle by engine from reservoir  ,
                $$\eta=\frac{W}{q_{1}}=\frac{Q_{1}-Q_{2}}{Q_{1}}=1-\frac{Q_{2}}{Q_{1}}$$
    if $$Q_{2}=0$$ i.e. heat is not given to sink , it means total heat absorbed is converted into work , then
                $$\eta=1$$ , which is practically not possible .
    Therefore $$Q_{2}\neq0$$ .
    Now as $$Q_{1}-Q_{2}=W$$ , therefore $$Q_{2}>W$$.

  • Question 10
    1 / -0
    Gas constant $$(R)$$ equals to
    Solution
    By Mayer's formula , we have 
              $$C_{P}-C_{V}=R$$
    Where $$C_{P}$$ and $$C_{V}$$ are the molar  specific heats of gas at constant pressure and constant volume respectively .
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