An ideal gas undergoes the thermodynamic cycle shown in the figure, where all tr

Question

An ideal gas undergoes the thermodynamic cycle shown in the figure, where all transformations are reversible. adiabatic 2 adiabatic 4 2 4 The pressure, volume and temperature of states i-1,, 4 are indicated by P, V, and Ti, respectively. For each of the three transformations composing the cycle, answer the following questions a) Is the work w done on the system positive, negative or zero? Is the exchanged heat q positive, negative or zero? Does the internal energy of the gas increase, decrease or stay constant? Does the entropy of the gas increase, decrease or stay constant? Does the temperature of the gas increase, decrease or stay constant? You don't need to carry out any calculation but you should briefly justify your answers. The sign convention for q and w is such that dU dq+dw.

Explanation / Answer

a)

1. work done by system is calculated by area under the curve,which is always positive.

2.heat is negative,as we have waste energy in sink after the whole process. i,e,energy is released i.e, negative

3. u=q+w

U= -ve + (+ve) depending on the magnitude of ,w & q we get U= +ve,(w>q)

U=-ve (w<q)

4. del S=del q/T

it is calculated by stpwise process, in isothermal which is not shown, t is const, q varies(-ve)

so, S =-ve(expansion)

in adiabatic process as shown, q is const, T varies (+ve)

so,S= +ve(compression)

5. in isothermal which is not shown, t is const,

   in adiabatic process as shown, q is const, T varies (+ve)

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