Question

A gas in a cylinder is held at a constant pressure of 1.80 $\times$ 10${}^5$ $Pa$ and is cooled and compressed from 1.70 m${}^3$ to 1.20 m${}^3$. The internal energy of the gas decreases by 1.40 $\times$ 10${}^5$ J. (a) Find the work done by the gas. (b) Find the absolute value of the heat flow, [$Q$] , into or out of the gas, and state the direction of the heat flow. (c) Does it matter whether the gas is ideal? Why or why not?

Short answer

(a) Work done by the gas is $-9.0\times {10}^4$ J. (b) Heat flow out of gas is $|Q|=2.30\times {10}^5$ J. (c) Gas type (ideal or real) does not matter in this scenario.

Step-by-step solution

Define the variables

We are given the initial and final volumes, the pressure, and the change in internal energy of the gas. Let the initial volume be $V_i=1.70{\text{m}}^3$, the final volume be $V_f=1.20{\text{m}}^3$, the pressure $P=1.80\times {10}^5\text{Pa}$, and the change in internal energy $\mathrm{\Delta }U=-1.40\times {10}^5\text{J}$.

Calculate the work done by the gas

The work done by the gas during compression can be calculated using the formula $W=P\times (V_f-V_i)$, where $W$ is the work done, $P$ is the pressure, and $V_f$ and $V_i$ are the final and initial volumes. Substituting the values, we get$$W=1.80\times {10}^5\text{Pa}\times (1.20-1.70){\text{m}}^3$$$$W=1.80\times {10}^5\text{Pa}\times (-0.50){\text{m}}^3$$$$W=-9.0\times {10}^4\text{J}$$

Use the first law of thermodynamics

According to the first law of thermodynamics, $\mathrm{\Delta }U=Q-W$, where $\mathrm{\Delta }U$ is the change in internal energy, $Q$ is the heat exchanged, and $W$ is the work done by the system. We have $\mathrm{\Delta }U=-1.40\times {10}^5\text{J}$ and $W=-9.0\times {10}^4\text{J}$. Rearrange to find $Q$:$$Q=\mathrm{\Delta }U+W$$$$Q=-1.40\times {10}^5\text{J}+(-9.0\times {10}^4\text{J})$$$$Q=-2.30\times {10}^5\text{J}$$

Determine the direction of heat flow

Since $Q=-2.30\times {10}^5\text{J}$, heat flows out of the gas as it is a negative value, indicating that the gas lost heat to the surroundings.

Analyze the need to consider the type of gas

For this problem, it does not matter whether the gas is ideal because we are dealing with macroscopic properties like pressure, volume, work, and energy that apply universally regardless of the microscopic model of the gas. The first law of thermodynamics holds true for any type of gas, whether ideal or real.

Key concepts

Work done by gas

When gas in a system undergoes volume change under constant pressure, it either does work or has work done on it. This concept is central to understanding how energy dynamics play out in thermodynamic systems.

In the exercise context, the gas is being compressed, which influences how we view 'work':

• The work done by the gas during compression is calculated with the formula: $$W=P\times (V_f-V_i)$$where $W$ is work, $P$ is pressure, and $V_f$, $V_i$ are final and initial volumes respectively.
• Here, since the final volume $V_f$ is smaller than the initial volume $V_i$, the value of $V_f-V_i$ is negative, indicating that work is done on the gas, not by the gas.
• Thus, the calculated work is negative: $$W=-9.0\times {10}^4\text{J}$$indicating energy being transferred into the gas, forcing it to compress.

Internal energy change

Internal energy is the total energy contained within a system due to its microscopic components. In thermodynamics, changes in internal energy provide insight into energy exchanges within a system.

The First Law of Thermodynamics, stating $\mathrm{\Delta }U=Q-W$, links internal energy change to heat flow and work:

• In this exercise, the problematic decrease in internal energy is $\mathrm{\Delta }U=-1.40\times {10}^5\text{J}$.
• Despite the work done on the gas, the total internal energy decreases, signifying that the energy output through heat flow exceeds the work input.
• Understanding how internal energy changes help depict the net result of energy transactions in a system, informing us about the thermodynamic path and resulting conditions of the system.

Heat flow direction

Heat flow plays a crucial role in understanding energy exchanges in thermodynamic processes. The direction and magnitude of heat flow tell us whether energy is entering or leaving the system.

Here's how this is applied in the exercise:

• Using the first law formula rearranged to solve for heat, $Q=\mathrm{\Delta }U+W$, informs us of the net heat transfer.$$Q=-1.40\times {10}^5\text{J}+(-9.0\times {10}^4\text{J})=-2.30\times {10}^5\text{J}$$
• The negative $Q$ value tells us heat is leaving the system, a critical sign of energy distribution as the gas cools and energy disperses into surroundings.
• This negative heat flow direction denotes that the thermal pathway counteracts internal energy changes, consistent with the physical context of a cooling gas under compression.