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Thermodynamics Test - 68

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Thermodynamics Test - 68
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  • Question 1
    1 / -0
    In changing the state of thermodynamics from A to B state, the heat required is Q and the work done by the system is W. The change in its internal energy is 
    Solution
    From the first law of thermodynamics
    $$\Delta Q=\Delta U+\Delta W $$ 
    $$W=+ve, Q=+ve$$
    $$ \Rightarrow \Delta U =\Delta Q - \Delta W $$
  • Question 2
    1 / -0
    First law of thermnodynamics is given by
    Solution
    The first law of thermodynamics is given as  $$\Delta Q=\Delta U+\Delta W $$ and $$\Delta W = P\Delta V $$, where $$\Delta U$$ is the change in internal energy of a system, $$Q$$ is the net heat transfer (the sum of all heat transfer into and out of the system), and $$W$$ is the net work done (the sum of all work done on or by the system)

  • Question 3
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    The first law of thermodynamics is concerned with the conservation of
    Solution
    The First Law of Thermodynamics states that heat is a form of energy, and thermodynamic processes are therefore subject to the principle of conservation of energy. This means that heat energy cannot be created or destroyed
  • Question 4
    1 / -0
    A system perform work $$\bigtriangleup W$$ when an amount of heat $$\bigtriangleup Q$$ added to the system the corresponding change in the internal energy is $$\bigtriangleup U$$. A unique function of the intial and final states (irrespective of the mode of change ) is
    Solution
    Change in internal energy $$ \left ( \Delta U  \right ) $$ depends upon initial an find state of the function while  $$ \Delta  Q $$  and $$ \Delta W $$ are path depends also. 
  • Question 5
    1 / -0
    If the amount of heat given to a system be 35 joules and the amount of work done by the system be - 15 joules, then the change in the internal energy of the system is
    Solution
    From the first law of thermodynamics.
    $$Q=+ve $$ because heat is given to the system and $$W=-ve$$ because work done is by the system.
    $$ \Delta Q = \Delta W +\Delta U\Rightarrow 35 = -15+\Delta U\Rightarrow \Delta U = 50 J $$
  • Question 6
    1 / -0
    For an idea gas, in an isothermal process
    Solution
    An isothermal process is a change of a system, in which the temperature remains constant
  • Question 7
    1 / -0
    In an isothermal expansion
    Solution
    In isothermal expansion temperature remains constant, hence no change in internal energy.
  • Question 8
    1 / -0
    In isothermic process, which statement is wrong
    Solution
    In isothermal process,  exchange of energy takes place between system and surrounding to maintain the system temperature constant.
  • Question 9
    1 / -0
    In a thermodynamic process, pressure of a fixed mass of a gas is changed in such a manner that the gas molecules given out $$30$$ J of heat and $$10$$ J of work is done on the gas. If the initial internal energy of the gas was $$40$$ J. Then the final internal energy will be
    Solution
    From the first law of Thermodynamics
    $$ \Delta Q=\Delta U+\Delta W = \left ( U_{f}-U_{f} \right )+\Delta W $$
    $$\Rightarrow 30=\left ( U_{f}-40 \right ) + 10\Rightarrow U_{f}= 60J $$

  • Question 10
    1 / -0
    A thermally container is divided into two parts by a screen. In one part the pressure and temperature are P and t for an idea gas filled. In the second part it is vaccum. If mow a small hole is created in the screen, then the temperature of the gas will
    Solution
    This is the case of free expansion of gas. In free expansion $$\Delta U=0\Rightarrow $$ Temp. remains same.
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