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First Law of Thermodynamics: $\Delta {\text{U}} = {\text{Q}} - {\text{W}}$

Where, $\Delta {\text{U}}$ is the internal energy of the given closed system

${\text{Q}}$ is the heat released or absorbed during the reversible isothermal expansion of the given gas.

And, ${\text{W}}$ is the work done during the reversible isothermal expansion of the given gas.

The work done in an isothermal reversible expansion is given by,

${\text{W}} = - nRT\ln \left( {\dfrac{{{V_2}}}{{{V_1}}}} \right) = - nRT\ln \left( {\dfrac{{{P_1}}}{{{P_2}}}} \right)$

Where, $n$ is the number of moles of the ideal gas present in the given system

${V_1}$ and ${V_2}$ are the initial and final volumes of the system during the process of expansion.

And, ${P_1}$ and ${P_2}$ are the initial and final pressures of the system during the process of the expansion

It is given that the process is isothermal, i.e., there is no change in the temperature of the system. Since the internal energy of a system depends on the temperature, hence, we can conclude that the change in internal energy of the system will be zero.

From the equation of the first law of thermodynamics, we get, $\Delta {\text{U}} = {\text{Q}} - {\text{W}}$

$ \Rightarrow {\text{Q}} = - {\text{W}}$

We know that the work done in an isothermal reversible expansion is given by

${\text{W}} = - nRT\ln \left( {\dfrac{{{P_1}}}{{{P_2}}}} \right)$

Substituting the values in the above equation, we get,

${\text{W}} = - \left( 1 \right) \times 8.314 \times 273 \times \ln \left( {\dfrac{1}{{0.1}}} \right)$

$ \Rightarrow {\text{W}} = - 8.314 \times 273 \times \ln 10$

Solving this:

$ \Rightarrow {\text{W}} = - 5226.23{\text{ J}}$

The first law of thermodynamics is an adapted form of the law of conservation of energy which has been modified so as to be applied in thermodynamic processes. It distinguishes two types of transfer of energy, as the heat transferred and the thermodynamic work done, during a process and relates them to Internal energy of the system which is a function of a system's state.

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