Question

A gas in a cylinder expands from a volume of 0.110 m\(^3\) to 0.320 m\(^3\). Heat flows into the gas just rapidly enough to keep the pressure constant at 1.65 \(\times\) 10\(^5\) Pa during the expansion. The total heat added is 1.15 \(\times\) 10\(^5\) J. (a) Find the work done by the gas. (b) Find the change in internal energy of the gas. (c) Does it matter whether the gas is ideal? Why or why not?

Short answer

(a) Work done: 3.465 × 10⁴ J. (b) Change in internal energy: 8.035 × 10⁴ J. (c) It doesn't matter if the gas is ideal.

Step-by-step solution

Understanding the Problem

We are given the initial and final volumes of a gas in a cylinder, the constant pressure during its expansion, and the total heat added to the system. Our goals are to find the work done by the gas, the change in its internal energy, and consider whether the ideal gas model affects the outcome.

Calculate the Work Done by the Gas (a)

The work done by the gas during an isobaric (constant pressure) process is given by the equation: \[ W = P \Delta V \]where \( P \) is the pressure and \( \Delta V \) is the change in volume. Given:\( P = 1.65 \times 10^5 \) Pa, initial volume \( V_i = 0.110 \) m\(^3\), final volume \( V_f = 0.320 \) m\(^3\).Calculate \( \Delta V \):\[ \Delta V = V_f - V_i = 0.320 - 0.110 = 0.210 \text{ m}^3 \]Substitute into the equation for work:\[ W = 1.65 \times 10^5 \times 0.210 = 3.465 \times 10^4 \text{ J} \]

Calculate the Change in Internal Energy (b)

The change in internal energy \( \Delta U \) for a system can be determined using the first law of thermodynamics:\[ \Delta U = Q - W \]where \( Q \) is the heat added to the system and \( W \) is the work done by the system.Given that \( Q = 1.15 \times 10^5 \) J (heat added) and we previously found \( W = 3.465 \times 10^4 \) J:\[ \Delta U = 1.15 \times 10^5 - 3.465 \times 10^4 = 8.035 \times 10^4 \text{ J} \]

Consideration of Gas Type (c)

In this problem, it does not matter whether the gas is ideal when calculating work done and internal energy change because these calculations depend solely on measurable quantities such as pressure, volume, and heat, not on the type of gas. The ideal gas assumption typically affects state equations but not directly the conservation of energy relations used here.

Key concepts

Isobaric Process

An isobaric process is a thermodynamic process that occurs at a constant pressure. This means that throughout the entire process, the pressure of the system does not change. In our exercise, the expansion of the gas happens isobarically at a steady pressure of 1.65 \( \times \) 10\(^5\) Pa.
Here are a few important points about isobaric processes:

• The volume of the gas changes, while the pressure remains constant.
• Work done by the gas in this process can be calculated using the formula \( W = P \Delta V \), where \( W \) is the work done, \( P \) is the constant pressure, and \( \Delta V \) is the change in volume.
• This process commonly occurs in everyday contexts, such as the heating of a gas in a sealed container where external pressure is constant.

Isobaric processes are straightforward because they simplify the relationship between pressure, volume, and the work done on or by a gas. The simplification arises due to constant pressure, making calculations easier to manage directly compared to processes where pressure varies.

First Law of Thermodynamics

The first law of thermodynamics is a form of the law of conservation of energy, adapted for thermodynamic systems. It establishes a relationship between internal energy, heat added to the system, and work done by the system. The first law states that the change in a system's internal energy \( \Delta U \) is equal to the heat added to the system \( Q \) minus the work done by the system \( W \). This can be expressed with the formula:
\[ \Delta U = Q - W \]
Where:

• \( \Delta U \) represents the change in internal energy.
• \( Q \) is the total heat added to the system.
• \( W \) is the work done by the system.

In our example, the heat added is 1.15 \( \times \) 10\(^5\) J,
and we calculated the work done by the gas as 3.465 \( \times \) 10\(^4\) J.
By utilizing the first law, we compute \( \Delta U \),
resulting in a significant shift in internal energy.
Understanding this principle is crucial because it enables us to analyze energy transfer in or out of any system, explaining how and why energy changes within the system.

Ideal Gas

In the context of thermodynamics, an 'ideal gas' is a hypothetical gas whose molecules occupy negligible space and have no interactions, except for elastic collisions. An ideal gas follows the ideal gas law, which is generally stated as \( PV = nRT \), where

• \( P \) is the pressure,
• \( V \) is the volume,
• \( n \) is the number of moles of the gas,
• \( R \) is the universal gas constant,
• \( T \) is the absolute temperature.

The consideration of a gas being ideal does not affect the calculations for the change in internal energy or the work done in our original problem, because these calculations rely only on measurable parameters such as pressure, volume, and heat. It's when we need to understand how gases behave on a micro-level, considering relationships like pressure, volume, and temperature, that the ideal gas model becomes significant.
While real gases deviate from the ideal gas model under high pressure and low temperature, for many practical purposes, the ideal gas law provides helpful approximations, simplifying calculations and predictions about gas behavior. In our exercise, discussing whether the gas is ideal is more relevant to theoretical interests rather than affecting practical outcomes in these specific calculations.