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Question: For an isothermal process: \(\begin{aligned} & \left( A \right)dQ=dW \\\ & \left( B \right...

For an isothermal process:
(A)dQ=dW (B)dQ=dU (C)dW=dU (D)dQ=dU+dW \begin{aligned} & \left( A \right)dQ=dW \\\ & \left( B \right)dQ=dU \\\ & \left( C \right)dW=dU \\\ & \left( D \right)dQ=dU+dW \\\ \end{aligned}

Explanation

Solution

The process in which temperature of the system remains constant is called isothermal process. Thermodynamic definition of heat is when a system and its surroundings are at different temperatures and on which work may undergo a process, the energy transferred by non-mechanical means is equal to the difference between the internal energy change and work done.

Formula used:
Q=(UfUi)WQ=\left( {{U}_{f}}-{{U}_{i}} \right)-W
where, Q is the heat
U is the internal energy
W is the work done

Complete answer:
The process in which the temperature of the system remains constant is called an isothermal process.
Thermodynamic definition of heat is when a system and its surroundings are at different temperatures and on which work may undergo a process, the energy transferred by non-mechanical means is equal to the difference between the internal energy change and work done.
Thus according to the first law of Thermodynamics,
dQ=dU+dWdQ=dU+dW
In an isothermal process the internal energy is equal to zero. The internal energy is a function of temperature.
That is, dU=0
dQ=dWdQ=dW

Thus option (A ) is correct.

Additional information:
According to ideal gas equation,
PV=nRTPV=nRT
Since T= a constant in an isothermal process the equation for isothermal process is,
PV= a constant=K
To calculate the work done in an isothermal process,
W=ifPdVW=-\int\limits_{i}^{f}{PdV}
W=ViVfPdVW=-\int\limits_{{{V}_{i}}}^{{{V}_{f}}}{PdV}
But by equation PV=K
Rearranging the equation we get,
P=KVP=\dfrac{K}{V}
W=ViVfKVdVW=-\int\limits_{{{V}_{i}}}^{{{V}_{f}}}{\dfrac{K}{V}dV}
W=K[lnVflnVi]W=-K\left[ \ln {{V}_{f}}-\ln {{V}_{i}} \right]
W=K[lnV]ViVfW=-K\left[ \ln V \right]_{{{V}_{i}}}^{{{V}_{f}}}
W=K[lnVfVi]W=-K\left[ \ln \dfrac{{{V}_{f}}}{{{V}_{i}}} \right]
W=KlnViVfW=K\ln \dfrac{{{V}_{i}}}{{{V}_{f}}}
Thus PfVf=PiVi=K{{P}_{f}}{{V}_{f}}={{P}_{i}}{{V}_{i}}=K
Hence, work done in an isothermal process is,
W=PfVflnViVf=PiVilnViVfW={{P}_{f}}{{V}_{f}}\ln \dfrac{{{V}_{i}}}{{{V}_{f}}}={{P}_{i}}{{V}_{i}}\ln \dfrac{{{V}_{i}}}{{{V}_{f}}}

Note:
The process in which the temperature of the system remains constant is called an isothermal process. In an isothermal process the internal energy is equal to zero. The internal energy is a function of temperature. Thermodynamic definition of heat is when a system and its surroundings are at different temperatures and on which work may undergo a process, the energy transferred by non-mechanical means is equal to the difference between the internal energy change and work done.