Question

A fluid undergoes a reversible adiabatic compression from 0.5 MPa, 0.2m^3 to 0.05m^3 according to the law PV^1.3=constant. Determine the change in enthalpy, internal energy and heat transfer and work transfer during the process.

Answer 1

Adiabatic process is a process in which a thermodynamic system does not exchange heat or substance with its surroundings. Hence, in the given adiabatic compression, the heat transfer $Q$ is equal to zero.

The first quantity we can determine is the work done by the system on its surroundings. It is calculated by taking the following integral:

$$W = \int_{V_1}^{V_2} p dV \, ,$$

where $p$ is pressure, and $V_1 = 0.2\, \text{m}^3$ and $V_2 = 0.05\, \text{m}^3$ is the initial and final volume, respectively. The pressure as a function of volume is provided by the conditions of the problem: $p V^{1.3} = \text{const} = p_1 V_1^{1.3}$, where $p_1 = 0.5\, \text{MPa} = 0.5 \times 10^6\, \text{N}/\text{m}^2$ is the initial pressure. Hence,

$$p = p_1 \left( \frac{V_1}{V} \right)^{1.3} \, .$$

Substituting this into the above integral and performing the elementary integration, we obtain

$$W = \frac{p_1 V_1}{0.3} \left[ 1 - \left( \frac{V_1}{V_2} \right)^{0.3} \right] \approx - 1.72 \times 10^5\, \text{J}\, ,$$

where we have taken into account that $\text{N} \cdot \text{m} = \text{J}$. Thus, the positive work is actually done by the surroundings on the system.

The change $\Delta U$ in the internal energy of the system is determined from the first law of thermodynamics:

$$\Delta U = Q - W = - W = 1.72 \times 10^5\, \text{J} \, ,$$

since $Q = 0$.

The enthalpy of a system is defined as $H = U + p V$. The change in enthalpy in the given process is, therefore,

$$\Delta H = \Delta U + p_2 V_2 - p_1 V_1 \, .$$

To find the product $p_2 V_2$, we use the equation $p_2 V_2^{1.3} = p_1 V_1^{1.3}$, whence $p_2 V_2 = p_1 V_1 \left( V_1 / V_2 \right)^{0.3}$. Eventually,

$$\Delta H = \Delta U + p_1 V_1 \left[ \left( \frac{V_1}{V_2} \right)^{0.3} - 1 \right] \\ {} = - W - 0.3\, W = - 1.3 \, W \approx 2.24 \times 10^5\, J \, .$$