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Question: How do you find heat in isothermal processes?...

How do you find heat in isothermal processes?

Explanation

Solution

The isothermal process is defined as the thermodynamic process in which the temperature of the system is being constant. The transfer of the heat into and out of the system takes place slowly so that the thermal equilibrium is maintained. The thermal is the word which describes the heat of the system.

Complete step-by-step answer: To approach this question let us first know the first law of thermodynamics. The first law of thermodynamic is :
ΔU=q+w\Delta U = q + w
Here the q is the heat flow, w represents the work and ΔU\Delta Urepresents the change in the internal energy.
In the isothermal process the internal energy is tend to be the function of the temperature.Now according to this the first law of the thermodynamic becomes:
w=−qw = - q
So for the reversible process when the efficiency is maximum, the formula becomes:
wrev=−∫V1V2PdV=−qrev{w_{rev}} = - \int\limits_{{V_1}}^{{V_2}} {PdV} = - {q_{rev}}
Here the work done is having a negative sign which means the work done is been done by the system onto the surrounding.
This further means that when the system of the gaseous particles gets expanded at a constant temperature the system ability to expand is due to the heat energy required. This means that the heat which has been flowing goes into the pressure volume work and does not tend to change the temperature.
Let us take an example in which 1 mole of the ideal gas has been expanded reversibly to double its volume at 298.15K.
Here the reversible work which has given rise to the expansion is found by using the ideal gas law for the pressure:
So here:
wrev=−∫V12V2nRTVdV{w_{rev}} = - \int\limits_{{V_1}}^{2{V_2}} {\dfrac{{nRT}}{V}dV}
On integrating we get,
wrev=−nRTln⁡(2V1V1)=−nRTln⁡2{w_{rev}} = - nRT\ln (\dfrac{{2{V_1}}}{{{V_1}}}) = - nRT\ln 2
Now substituting the values we get,
wrev=−1×8.314472×298.15ln⁡2{w_{rev}} = - 1 \times 8.314472 \times 298.15\ln 2
On solving we get,
wrev=−1718.28J{w_{rev}} = - 1718.28J
Now the heat which would be flowing for performing the expansion will be:
qrev=−wrev{q_{rev}} = - {w_{rev}}
On substituting the value we get,
qrev=+1718.28J{q_{rev}} = + 1718.28J
So the answer will be qrev=+1718.28J{q_{rev}} = + 1718.28J

Note: The change in the temperature in the isothermal process is zero. So due to which the change in the internal energy would be zero. The isothermal process is slow because it works at a constant temperature so that the temperature gets maintained.