Question

A gas expands in a piston-cylinder assembly from $p_1=8$ bar, $V_1=0.02{\mathrm{m}}^3$ to $p_2=2$ bar in a process during which the relation between pressure and volume is $p{V}^{1.2}=$ constant. The mass of the gas is $0.25\mathrm{kg}$. If the specific internal energy of the gas decreases by $55\mathrm{kJ}/\mathrm{kg}$ during the process, determine the heat transfer, in kJ. Kinetic and potential energy effects are negligible.

Short answer

The heat transfer is -23.08 kJ.

Step-by-step solution

- Identify given information

Given:Initial pressure, $p_1=8$ bar = $800$ kPaInitial volume, $V_1=0.02$ m³Final pressure, $p_2=2$ bar = $200$ kPaMass of gas, $m=0.25$ kgDecrease in specific internal energy, $\mathrm{△}u=55$ kJ/kg

- Determine the final volume

Given the relation $p{V}^{1.2}=$ constant, use the relation with initial and final conditions:p_1 V_1^{1.2} = p_2 V_2^{1.2}$800kPa\times (0.02{)}^{1.2}=200kPa\times V_2^{1.2}$Solve for $V_2$:$V_2=\sqrt[1.2]{\frac{800\times (0.02{)}^{1.2}}{200}}=0.04664$ m³

- Calculate work done by the gas

Work done by the gas in a polytropic process is given by the formula:$$W=\frac{p_2{V}_2-p_1{V}_1}{1-n}$$where $n=1.2$Substitute the values obtained:$$W=\frac{200\times 0.04664-800\times 0.02}{1-1.2}=-9.33$$ kJ

- Use first law of thermodynamics

The first law of thermodynamics is expressed as:\Delta U = Q - W\where $\mathrm{\Delta }U=m\times \mathrm{△}u=0.25\times (-55)=-13.75$ kJRearranging the first law equation to solve for $Q$ (heat transfer):$$Q=\mathrm{\Delta }U+W$$Substitute the values of $\mathrm{\Delta }U$ and $W$:$$Q=(-13.75)+(-9.33)=-23.08$$ kJ

Key concepts

Polytropic Process

A polytropic process is a thermodynamic process that follows the equation: $$p{V}^n=\text{constant}$$ where $p$ is the pressure, $V$ is the volume, and $n$ is the polytropic index. This index can vary depending on the specific characteristics of the gas and the process. In this exercise, n is given as 1.2. Understanding the polytropic index is essential because it dictates how the pressure and volume of the gas interrelate during the process. For example, if $n=1$, the process is isothermal, meaning the temperature remains constant. If $n=\text{Cp/Cv}$ (for an ideal gas), the process is adiabatic, meaning no heat is transferred.

First Law of Thermodynamics

The first law of thermodynamics is a statement of energy conservation. It states that the energy added to a system as heat ($Q$) minus the energy taken out of the system as work ($W$) equals the change in the system's internal energy ($\mathrm{△}U$). The equation can be written as: $$\mathrm{△}U=Q-W$$ This principle is crucial in solving thermodynamics problems because it provides a relationship between heat transfer, work done, and changes in internal energy. In our problem, we need to rearrange this formula to solve for the heat transfer ($Q$) after finding the work done ($W$) and the change in internal energy ($\mathrm{△}U$).

Work Done by Gas

The work done by a gas during a polytropic process is derived from the pressure-volume relationship. The general formula for a polytropic process is: $$W=\frac{p_2{V}_2-p_1{V}_1}{1-n}$$ Here, $p_1$ and $p_2$ are the initial and final pressures, $V_1$ and $V_2$ are the initial and final volumes, and $n$ is the polytropic index. To find the work done, you need to know both the initial and final states of the system. For this exercise, substituting the given values into the formula allows us to compute the work done by the gas. Negative work indicates that work was done on the surroundings by the gas.

Pressure-Volume Relationship

In thermodynamics, the pressure-volume relationship is key to understanding how a system behaves as it undergoes different processes. For a polytropic process like the one in this exercise, the relationship is given by: $$p{V}^{1.2}=\text{constant}$$ This relationship helps to determine missing variables, such as the final volume or pressure, if other variables are known. In our solution, using the given initial and final pressures along with the initial volume, we solved for the final volume. This type of relationship is indispensable for solving many types of thermodynamic problems.