Question

A cylinder with a sliding piston contains an ideal gas. The cylinder (with the gas inside) is heated by transferring the same amount of heat to the system in two different ways: (1) The piston is blocked to prevent it from moving. (2) The piston is allowed to slide frictionlessly to the end of the cylinder. How does the final temperature of the gas under condition 1 compare to the final temperature of the gas under condition 2 ? Is it possible for the temperatures to be equal?

Short answer

Explain your answer. Answer: No, it is not possible for the final temperatures to be equal in both cases. The final temperature of the gas with the piston blocked is higher than the final temperature of the gas with the sliding piston, as the heat is also used to do work in the second case.

Step-by-step solution

Understand the problem and identify the given information

We have a cylinder containing an ideal gas. The same amount of heat (\(Q\)) has been added to the system in two different ways: (1) With the piston blocked, and (2) With the sliding piston. We want to compare the final temperatures of the gas in both cases and determine whether it is possible for the temperatures to be equal.

Use the first law of thermodynamics

According to the first law of thermodynamics, the heat added to a closed system is equal to the change in internal energy plus the work done by the system: \(Q = ΔU + W\) For an ideal gas, internal energy only depends on temperature, so \(ΔU = nC_VΔT\), where \(n\) is the number of moles, \(C_V\) is the molar heat capacity at constant volume, and \(ΔT\) is the change in temperature.

Analyze case 1 - The piston is blocked

Since the piston is blocked, the volume remains constant and no work is done by the gas. Thus, \(W = 0\). Substituting the expression for the change in internal energy from step 2, we have: \(Q = nC_VΔT_1\) Dividing by \(nC_V\), we find the change in temperature: \(ΔT_1 = \frac{Q}{nC_V}\)

Analyze case 2 - The piston slides frictionlessly

Since the piston can slide, the volume can change and work is done by the gas. For an ideal gas, the work done is equal to \(W = nRΔT_2\), where \(R\) is the universal gas constant, and \(ΔT_2\) is the change in temperature in this case. Substituting the expression for the change in internal energy from step 2, we have: \(Q = nC_VΔT_2 + nRΔT_2\) Dividing by \(n(C_V + R)\), we find the change in temperature: \(ΔT_2 = \frac{Q}{n(C_V + R)}\)

Compare the final temperatures

Comparing the expressions for \(ΔT_1\) and \(ΔT_2\), we see that: \(ΔT_1= \frac{Q}{nC_V}\) \(ΔT_2 = \frac{Q}{n(C_V + R)}\) Since \(C_V > 0\) and \(R > 0\), it follows that \(C_V < (C_V + R)\), which implies that: \(ΔT_1 > ΔT_2\) The final temperature of the gas under condition 1 (piston blocked) is higher than the final temperature of the gas under condition 2 (piston sliding).

Evaluate the possibility of equal temperatures

For the final temperatures to be equal, we must have: \(ΔT_1 = ΔT_2\) This would imply: \(\frac{Q}{nC_V} = \frac{Q}{n(C_V + R)}\) Since \(Q > 0\) and \(n > 0\), it would require \(C_V = (C_V + R)\) to satisfy this equation. However, this is not possible, since \(R > 0\). Thus, the final temperatures of the gas under conditions 1 and 2 cannot be equal.

Key concepts

First Law of Thermodynamics

Understanding the first law of thermodynamics is crucial for anyone studying the behavior of gases and heat transfer. This fundamental law states that energy can neither be created nor destroyed, only transformed from one form to another. In the context of an ideal gas in a cylinder, this means that when we add heat (\(Q\)), that energy must go into either increasing the gas's internal energy (\(ΔU\)), which is related to its temperature, or doing work (\(W\)) on the surroundings — such as moving a piston.

If the piston is fixed, all the heat added goes into raising the internal energy, which directly increases the gas's temperature. Conversely, if the piston is allowed to move, some of the added heat does work by moving the piston, meaning not all of it contributes to the temperature increase. This distinction is at the core of understanding why the final temperatures vary in our exercise under different conditions.

Ideal Gas

An ideal gas is a theoretical construct, a simplified model that helps us predict the behavior of real gases under various conditions. The model assumes that the gas particles are point masses without volume and that there are no intermolecular forces between them — they collide elastically without losing energy.

The ideal gas law, given by \(PV = nRT\), where \(P\) stands for pressure, \(V\) for volume, \(n\) for number of moles, \(R\) for the ideal gas constant, and \(T\) for temperature, is a key equation in understanding how temperature, pressure, and volume relate for an ideal gas. When applying this model to the problem at hand, it helps explain the work done when the piston moves (case 2), which is not present when the piston is blocked (case 1).

Heat Capacity

Heat capacity is a measure of the amount of heat energy required to raise the temperature of a substance. For a gas, we distinguish between the molar heat capacity at constant volume (\(C_V\)) and at constant pressure (\(C_P\)). The former is particularly relevant when dealing with a fixed volume, as in our blocked piston scenario. The molar heat capacity tells us how much heat is needed to raise the temperature of one mole of a substance by one degree Celsius at constant volume.

\(C_V\) is part of the equation \(Q = nC_VΔT\) for a process at constant volume. In contrast, \(C_P\) would be used for a process at constant pressure — though it's not directly part of this exercise, it's worth noting that \(C_P = C_V + R\) for an ideal gas. Understanding these capacities and their role in heat transfer is essential for predicting the outcome of our aforementioned heat transfer scenarios.