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Question: Look at the indicator diagram and answer the following questions: (i) Which thermodynamic process ...

Look at the indicator diagram and answer the following questions:
(i) Which thermodynamic process does it possibly represent?
(ii) How much work is done when the volume of the working substance changes from V1{V_1} to V2{V_2}?

Explanation

Solution

In this question, we need to comment on the thermodynamic process involved in the given figure and evaluate the work done when the volume of the working substance changes from V1{V_1} to V2{V_2}. For this, we will observe the given plot carefully and try to relate the different thermodynamic process with the given figure.

Complete step by step answer:
The graph of two variables xx and yy will be a rectangular hyperbola if the product x.yx.y is always constant. It is known that the temperature is constant in an isothermal thermodynamic process.

In the above figure, the graph plotted between the pressure PP and the volume VV is in hyperbolic nature. Which means, the product P.VP.V is constant. From the formula P.V=nRTP.V = nRT, we can say that the right hand side of the equation, nRTnRT is constant as the left hand side is also a constant. As the number of moles nn and the universal gas constant RR which is equal to 8.314J8.314J/molmol are both constants, it can be said that the temperature TT is also constant. Hence, the above indicator diagram represents an isothermal thermodynamic process.

Consider that in an isothermal process, the pressure and volume of the ideal gas changes from (P1,V1)({P_1},{V_1}) to (P2,V2)({P_2},{V_2}) . Somewhere in the middle of the process where the pressure is PP and volume changes from VV to V+ΔVV + \Delta V, ΔV\Delta V being infinitesimally small. From the first law of thermodynamics ΔW=PΔV\Delta W = P\Delta V, ΔV0\Delta V \to 0 as it is infinitesimally small and summing ΔW\Delta W over entire process, we get total work done by the gas as
W=V1V2PdVW = \int\limits_{{V_1}}^{{V_2}} {PdV} where limits of integration goes from V1{V_1} to V2{V_2}. Using the formula PV=nRTPV = nRT, we get P=nRTVP = \dfrac{{nRT}}{V}
W=V1V2(nRTV)dV\Rightarrow W = \int\limits_{{V_1}}^{{V_2}} {\left( {\dfrac{{nRT}}{V}} \right)dV}
On integrating the above equation, we get
W=nRTln(V2V1)W = nRT\ln \left( {\dfrac{{{V_2}}}{{{V_1}}}} \right) and as we know that ln=loge\ln = {\log _e}, this can be written as
W=2.303nRTlog(V2V1)W = 2.303nRT\log \left( {\dfrac{{{V_2}}}{{{V_1}}}} \right)

Note: The formula PV=nRTPV = nRT is used where PP is pressure, VV is volume, nn is the number of moles, RR is the universal gas constant and TT is the temperature. First law of thermodynamics: ΔW=PΔV\Delta W = P\Delta V where ΔW\Delta W is the change in work done, PP is the pressure and ΔV\Delta V is the change in volume. The work done can also be expressed in terms of initial pressure P1{P_1} and final pressure P2{P_2} as W=2.303nRTlog(P1P2)W = 2.303nRT\log \left( {\dfrac{{{P_1}}}{{{P_2}}}} \right).