A cylinder of oxygen ( considered as an ideal gas undergoes a process. If the ga

Question

A cylinder of oxygen ( considered as an ideal gas undergoes a process. If the gas performs 90 J of work and absorbs 160 J of heat. the change of its internal energy is ?
a) 70J b) 0J c) -70J

If the gas is maintained at a constant temperature and the gas expands to two time its original volume at the end of this process its pressure at the end is ?
a) is 2 times as large b) remains unchanged c) is one half of the original pressure

If the process is an isothermal process we have the following equation: ( U is the internal energy, W is the work done, and Q is the heat absorbed.) ?
a) Q = 0 b) W = 0 c) deltaU = 0 d) W = p deltaV

If the gas is maintained at constant pressure of 8.0 x 10^4 Pa and expands from 0.01 m^3 to 0.015 m^3, the work done by the gas is?
a) 800J b)1200J c) 400J

Explanation / Answer

The work doen by gas W = 90J

The heat absorded by gas Q = 160J

Then from first law of thermodynamics

        U = Q - W

             = 160 - 90 = 70J

From ideal gas equation

  Let the initial volume of the gas be Vi = V

and the final volume of the gas is Vf = 2V

Let the initial pressure of the gas Pi = P

then PiVi = PfVf

Therefore the pressure

      Pf = PiVi / Vf

          = Pi (V/2V)

         = P/2

Therefore the pressure becomes half of its orginal

When the process is isothermal, then there is no change in internal energy since the temperature remains constant, the internal energy remains constant

               so U = 0

The pressure P = 8.0 x 10^4 Pa

The change in volume V = 0.015-0.01 = 0.005 m^3

Therefore the work done

             W = Pdv

                 = (8.0*10^4)(0.005)

                 = 400J

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