Question

If 4.00 mol hydrogen $(C_p=28.8{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1})$is expanded reversibly and isothermally at 400 K from an initial volume of 12.0 L to a final volume of 30.0 L, calculate$\mathbf{\Delta U},\mathbf{q}\mathbf{,}\mathbf{w}\mathbf{,}\mathbf{\Delta H}\mathbf{,}\mathbf{and} \mathbf{\Delta S}$ for the gas.

Short answer

The thermodynamic parameters are:
$$\begin{gathered} w=-12.2k\mathit{}J \\ q=12.2 k\mathit{}J \\ \Delta H=\Delta U=0 \\ \Delta S=30.5J{K}^{-1} \end{gathered}$$

Step-by-step solution

Heat energy

Heat energy is a measure of the total internal energy of a system. It is found by combining the total kinetic and potential energy of the molecule.

The concept.

The internal energy U depends only on temperature and for the isothermal process, the temperature is constant, and the change in internal energy$∆\mathrm{U}$ will be zero.
$\Delta U=0$

The work done for the isothermal reversible process is given by the formula
$w=-nRT\ln \frac{V_2}{V_1}$

*n is the moles of gas, i.e., 4.00 mol.

*R is the universal gas constant, i.e.,$8.314J{K}^{-1}mo{l}^{-1}$

*T is the absolute temperature, i.e., 400 K.

*V₂ is the final temperature, i.e., 30.0 L.

*V₁ is the initial temperature, i.e., 12.0 L.

The Work Done.

Substituting the known values, in the above equation.
$$\begin{gathered} w=-4.00mol\times 8.314J{K}^{-1}mo{l}^{-1}\times 400K\times \ln \frac{30.0L}{12.0 L} \\ =-13302.4J\times 0.916 \\ =-12.18\times {10}^{3}J \\ =12.2 kJ \end{gathered}$$

The Heat.

For the isothermal process ,$\Delta U=0$so the amount of heat involved is calculated from the first law of thermodynamics as:
$$\begin{gathered} \Delta U=q+w \\ q=\Delta U-w \\ =0-\left(-12.2kJ\right) \\ =12.2kJ \end{gathered}$$

The Enthalpy.

For isothermal process the change in enthalpy is calculated as:
$$\begin{gathered} \Delta H=\Delta U+\Delta \left(PV\right) \\ =\Delta U+nR\Delta T \\ =0+0 \\ =0 \end{gathered}$$
Thus, for isothermal process change in enthalpy is 0.

The Entropy.

At constant temperature, the entropy change is given by the equation as:
$$\begin{gathered} \Delta S=nR\ln \frac{V_2}{V_1} \\ =4.00mol\times 8.314J{K}^{-1}mo{l}^{-1}\times \ln \frac{30.0 L}{12.0L} \\ =33.2\times 0.916 \\ =30.5 J{K}^{-1} \end{gathered}$$
Thus, thethermodynamic parameters are:
$$\begin{gathered} \mathit{}\mathit{}w=-12.2kJ \\ q=12.2 kJ \\ \Delta H=\Delta U=0 \\ \Delta S=30.5J{K}^{-1} \end{gathered}$$