Problem 9 - University Physics with Modern Physics

Question

A gas in a cylinder expands from a volume of 0.110 \(\mathrm{m}^{3}\) to 0.320 \(\mathrm{m}^{3} .\) 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 \(1.15 \times 10^{5} \mathrm{J}\) (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?

Step-by-Step Solution

Answer

(a) 3.465 x 10^4 J; (b) 8.035 x 10^4 J; (c) No, it doesn't matter for these calculations.

Step 1: Identify the Known Values

First, we identify the given values from the problem. The initial volume \( V_i \) is 0.110 \( \mathrm{m}^3 \), the final volume \( V_f \) is 0.320 \( \mathrm{m}^3 \), the pressure \( P \) is constant at \( 1.65 \times 10^5 \) Pa, and the heat added \( Q \) is \( 1.15 \times 10^5 \) J.

Step 2: Calculate the Work Done by the Gas

The work done by a gas at constant pressure during expansion can be calculated using the formula: \( W = P \times \Delta V \), where \( \Delta V = V_f - V_i \). Substituting in the given quantities, we have \( \Delta V = 0.320 - 0.110 = 0.210 \) \( \mathrm{m}^3 \). Therefore, \( W = 1.65 \times 10^5 \) Pa \( \times 0.210 \mathrm{m}^3 = 3.465 \times 10^4 \) J.

Step 3: Calculate the Change in Internal Energy

To find the change in internal energy \( \Delta U \), use the first law of thermodynamics: \( \Delta U = Q - W \). Substituting the given heat and calculated work, \( \Delta U = 1.15 \times 10^5 \) J \(- 3.465 \times 10^4 \) J = \( 8.035 \times 10^4 \) J.

Step 4: Consider If Gas Being Ideal Matters

For part (c), consider whether it matters if the gas is ideal. The calculation of work done and the change in internal energy use thermodynamic principles (first law of thermodynamics) that are valid for both ideal and real gases. Whether the gas is ideal does not affect the mathematical results, but it may influence phenomena not covered by these calculations (like intermolecular forces or non-linear paths).

Key Concepts

Constant Pressure in Thermodynamics

In thermodynamics, when we talk about constant pressure, it refers to a process where the pressure of the system remains unchanged during operations like heating, cooling, compression, or expansion.
This kind of process is also known as isobaric.During an isobaric process:

• The pressure remains constant throughout.
• The system can still undergo changes in volume or temperature.
• It is often represented in a PV graph as a horizontal line.

The given exercise involves gas expanding in a cylinder at constant pressure of \(1.65 \times 10^5\) Pa. Keeping pressure constant is crucial as it influences how substances exchange heat, work, and internal energy.Understanding why pressure is kept constant helps recognize the relationship between pressure and volume changes described by Charles' law.
This is essential for calculations like work done in isobaric processes.

First Law of Thermodynamics

The first law of thermodynamics, a core principle of energy conservation, states that the energy in a closed system remains constant.
It is often expressed as:\[\Delta U = Q - W\]Where:

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

In our exercise, this principle is key for calculating the change in internal energy after the gas expands. The given problem specifies that there is a heat addition of \(1.15 \times 10^5\) J and requires calculating the work done. Using the first law, it is determined that \(\Delta U = 8.035 \times 10^4\) J.
This shows how energy transfers within the system, either increasing internal energy or doing work.

Work Done by Gas at Constant Pressure

The work done by gas during an expansion or compression at constant pressure is given by the formula:\[W = P \times \Delta V\]Where:

• \(W\) is the work done by the gas.
• \(P\) is the constant pressure.
• \(\Delta V = V_f - V_i\), the change in volume.

In our problem, the gas expands from an initial volume of 0.110 \( \mathrm{m}^3 \) to a final volume of 0.320 \( \mathrm{m}^3 \). The pressure remains constant at \(1.65 \times 10^5\) Pa, thus allowing us to calculate the work done as \(3.465 \times 10^4\) J.
This calculation is essential for understanding the energy transactions in an isobaric process. Grasping how to find the work done by gas ensures that students can apply this knowledge practically in various real-world situations.