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One mole of an ideal gas is heated slowly so that it goes from the \(P V\) state \(\left(P_{i}, V_{i}\right)\) to \(\left(3 P_{i}, 3 V_{i}\right)\) in such a way that the pressure is directly proportional to the volume. (a) How much work is done on the gas in the process? (b) How is the temperature of the gas related to its volume during this process?

Short Answer

Expert verified
The work done on the gas is equal to \(3P_{i}V_{i}\) and the temperature is directly related to the volume of the gas in this process.

Step by step solution

01

Identify the Relationship

Given that the pressure is directly proportional to the volume, this implies that \(P = kV\), where k is the constant of proportionality.
02

Apply the Ideal Gas Law

The ideal gas law states that \(PV = nRT\), where n is the number of moles, R is the gas constant, and T is the temperature. Since \(P = kV\), the ideal gas law can be rewritten as \(kV^2 = nRT\). Let's solve for T to get \(T = kV^2 / nR\).
03

Calculate Work Done

The work done on the gas is given by the integral of pressure over volume, i.e., \(W = \int_{V_{i}}^{V_{f}}{PdV}\). Substituting for \(P = kV\) we get \(W = \int_{V_{i}}^{3V_{i}}{kVdV} = k \left[ V^2 / 2 \right]_{V_{i}}^{3V_{i}} = 4kV_{i}^2 - kV_{i}^2 = 3kV_{i}^2\). Substituting \(P_{i} = kV_{i}\), we find that \(W = 3P_{i}V_{i}\).
04

Relate Temperature to Volume

From Step 2, we know, \(T = kV^2 / nR\). Which can be written as \(T = PV / nR\). Since in the given process, \(P = kV\), so \(T = kV^2 /nR = PV / nR\). This implies that the temperature is directly related to volume in this process.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Work Done on Gas
When talking about the work done on a gas, we're diving into how energy is transferred during expansion or compression. In this exercise, we saw that the gas expanded from an initial state \(V_i\) to a final state \(3V_i\) and a pressure change from \(P_i\) to \(3P_i\). This expansion is done in such a manner that pressure is directly proportional to volume.

To determine the work done, we use the equation:
  • \( W = \int_{V_{i}}^{3V_{i}} P \, dV \)
Given this proportionality (\(P = kV\)), when we integrate from \(V_i\) to \(3V_i\), we find the amount of work done.

Inserting the relationship into the integral, we perform the calculation:
  • \( W = \int_{V_{i}}^{3V_{i}} kV \, dV = k \left[ \frac{V^2}{2} \right]_{V_i}^{3V_i} = 3kV_i^2 = 3P_iV_i \)
This reflects the energy used (or work done) in expanding the gas within these constraints. The work done is the area under the curve in a Pressure-Volume graph, equivalent to energy transferred.
Temperature-Volume Relationship
The temperature-volume relationship in this context is central to our understanding of gas behavior under the conditions set by the problem. By using the Ideal Gas Law, \(PV = nRT\), we explore how temperature varies with volume.

Given that pressure is proportional to volume (\(P = kV\)), substituting into the ideal gas equation gives:
  • \(kV^2 = nRT\)
  • Reorganizing for temperature gives: \(T = \frac{kV^2}{nR}\)
This formula tells us that temperature is a function of the square of the volume, illustrating how temperature changes with respect to volume.

During the process described, as the gas volume increases from \(V_i\) to \(3V_i\), the temperature also varies accordingly due to this quadratic relationship. This means that temperature changes in a nonlinear manner with volume, which is key to predicting gas behavior under such conditions.
Proportional Pressure-Volume
Direct proportionality between pressure and volume, \(P = kV\), is a unique scenario within gas dynamics. Often we expect pressure to decrease as volume increases, but here, they increase together.

In this exercise, we see that as the gas expands from \(V_i\) to \(3V_i\), the pressure transitions similarly from \(P_i\) to \(3P_i\). This steady proportional growth is indicative of a controlled environment where energy input is constant throughout the process.
  • Understanding this relationship is key: as the volume triples, so too does the pressure.
  • This becomes crucial in systems where predictable pressure-volume behavior is needed, such as in hydraulics or controlled laboratory settings.
The beauty of this term in the Ideal Gas Law context is its simplification of calculations related to work done and temperature changes, allowing us deeper insights with more straightforward computation.

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Most popular questions from this chapter

A \(3.00-\mathrm{g}\) copper penny at \(25.0^{\circ} \mathrm{C}\) drops \(50.0 \mathrm{m}\) to the ground. (a) Assuming that \(60.0 \%\) of the change in potential energy of the penny-Earth system goes into increasing the internal energy of the penny, determine its final temperature. (b) What If? Does the result depend on the mass of the penny? Explain.

A gas is compressed at a constant pressure of 0.800 atm from \(9.00 \mathrm{L}\) to \(2.00 \mathrm{L}\). In the process, \(400 \mathrm{J}\) of energy leaves the gas by heat. (a) What is the work done on the gas? (b) What is the change in its internal energy?

The surface of the Sun has a temperature of about \(5800 \mathrm{K}\). The radius of the Sun is \(6.96 \times 10^{8} \mathrm{m} .\) Calculate the total energy radiated by the Sun each second. Assume that the emissivity of the Sun is 0.965.

A \(1.00-\mathrm{kg}\) block of copper at \(20.0^{\circ} \mathrm{C}\) is dropped into a large vessel of liquid nitrogen at \(77.3 \mathrm{K}\). How many kilograms of nitrogen boil away by the time the copper reaches \(77.3 \mathrm{K} ?\) (The specific heat of copper is \(0.0920 \mathrm{cal} / \mathrm{g} \cdot^{\circ} \mathrm{C}\). The latent heat of vaporization of nitrogen is 48.0 cal/g.)

How much work is done on the steam when 1.00 mol of water at \(100^{\circ} \mathrm{C}\) boils and becomes \(1.00 \mathrm{mol}\) of steam at \(100^{\circ} \mathrm{C}\) at 1.00 atm pressure? Assuming the steam to behave as an ideal gas, determine the change in internal energy of the material as it vaporizes.

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