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One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of \(27^{\circ} \mathrm{C}\). If work done during the process is \(3 \mathrm{~kJ}\), then final temperature of the gas is \(\left(\mathrm{C}_{\mathrm{v}}=20 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)\) (a) \(150 \mathrm{~K}\) (b) \(200 \mathrm{~K}\) (c) \(175 \mathrm{~K}\) (d) \(225 \mathrm{~K}\)

Short Answer

Expert verified
The final temperature of the gas is 150 K (option a).

Step by step solution

01

Convert Initial Temperature

The initial temperature is given as \(27^{\circ} \mathrm{C}\). To convert this to Kelvin, use the formula:\[ T_{\text{K}} = T_{\text{C}} + 273.15 \]Substitute the value:\[ T_{1} = 27 + 273.15 = 300.15 \mathrm{~K} \] For ease of calculation, we use \(T_{1} = 300 \mathrm{~K}\) (as effects of precision will not change the selections for an answer in this context).
02

Use First Law of Thermodynamics for Adiabatic Process

For an adiabatic process, the change in internal energy \((\Delta U)\) equals the work done \((W)\) since no heat is exchanged \((q = 0)\). Thus, we have:\[ \Delta U = -W \]Given \( W = 3 \mathrm{~kJ} \), or \(3000 \mathrm{~J}\):\[ \Delta U = -3000 \mathrm{~J} \]
03

Calculate Change in Temperature

The change in internal energy for one mole of an ideal gas is given by:\[ \Delta U = n \cdot C_{\text{v}} \cdot \Delta T \]where \( n = 1 \) mole, \( C_{\text{v}} = 20 \mathrm{~J/K~mol} \), and \( \Delta T = T_{2} - T_{1} \).Substitute the values:\[ -3000 = 1 \times 20 \times (T_{2} - 300) \]Simplifying, we get:\[ -3000 = 20(T_{2} - 300) \]
04

Solve for Final Temperature

Continuing the equation from the previous step:\[ -3000 = 20(T_{2} - 300) \]Divide each side by 20:\[ -150 = T_{2} - 300 \]Add 300 to each side to isolate \( T_{2} \):\[ T_{2} = 150 \mathrm{~K} \]
05

Confirm Answer

The calculated final temperature \( T_{2} \) is \( 150 \mathrm{~K} \), which matches option (a). Double-check calculations to ensure there have been no errors.

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Key Concepts

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

Ideal Gas
An ideal gas is a theoretical concept used in physics to simplify the behavior of gases. It is based on a set of assumptions about how gas particles interact. Within this model, it's assumed that the gas consists of a large number of small particles (molecules or atoms), all of which are in constant random motion. Additionally, ideal gas particles are not subject to intermolecular forces and the volume of the particles themselves is negligible compared to the container's volume.

There are several equations that describe the behavior of an ideal gas, the most commonly known being the Ideal Gas Law: \[ PV = nRT \]
  • \(P\) is the pressure of the gas,
  • \(V\) is the volume of the gas,
  • \(n\) is the amount of gas in moles,
  • \(R\) is the ideal gas constant, and
  • \(T\) is the temperature in Kelvin.

This law is often used as a starting point in various calculations that involve gases under moderate conditions.
First Law of Thermodynamics
The First Law of Thermodynamics is a fundamental principle that asserts the conservation of energy. It states that energy cannot be created or destroyed in an isolated system. Instead, it can only change forms. In the context of thermodynamics, the first law is often expressed as:\[ \Delta U = q - W \]where:
  • \( \Delta U \) is the change in internal energy of the system,
  • \( q \) is the heat absorbed by the system, and
  • \( W \) is the work done by the system.

In adiabatic processes, such as the one described in the exercise, there is no exchange of heat with the surroundings, so \( q = 0 \). Consequently, the change in internal energy is solely due to the work done: \( \Delta U = -W \). This is particularly useful in calculating changes in an adiabatic process.
Adiabatic Expansion
Adiabatic expansion occurs when a gas expands without any heat exchange with its surroundings. During this process, the system is insulated such that the heat \( q \) remains zero. The internal energy of the gas decreases due to the work done by the gas during expansion.

For one mole of an ideal gas, the relationship between the initial and final temperatures \( T_1 \) and \( T_2 \), when expanded adiabatically, can be determined through the internal energy equation:\[ \Delta U = nC_{v}(T_2 - T_1) = -W \]where
  • \( n \) is the number of moles,
  • \( C_v \) is the molar heat capacity at constant volume, and
  • \( W \) is the work done during the expansion.

This concept is vital within thermodynamics and helps explain various phenomena such as the temperature differences in an operating internal combustion engine.
Temperature Conversion
Temperature conversion is fundamental in physics and chemistry, particularly when dealing with gas laws and thermodynamic calculations. The exercise starts by converting the temperature from Celsius (°C) to Kelvin (K). This is necessary because temperature in physics often needs to be in absolute units (Kelvin) to comply with thermodynamic laws and equations.

The formula to convert from Celsius to Kelvin is straightforward:\[ T_{K} = T_{C} + 273.15 \]This relationship ensures that we avoid negative temperatures in calculations, as absolute zero (0 K) is considered the lowest limit for temperature.

Having all temperatures in Kelvin allows for consistency in calculations and better integration into formulas such as the Ideal Gas Law and other thermodynamic equations. Thus, effective temperature conversion ensures the reliability and accuracy of scientific calculations.

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Most popular questions from this chapter

Standard entropy of \(\mathrm{X}_{2}, \mathrm{Y}_{2}\) and \(\mathrm{XY}_{3}\) are 60,40 and 50 \(\mathrm{JK}^{-1} \mathrm{~mol}^{-1}\), respectively. For the reaction, \(1 / 2 \mathrm{X}_{2}+3 / 2 \mathrm{Y}_{2} \longrightarrow \mathrm{XY}_{3}, \Delta \mathrm{H}=-30 \mathrm{~kJ}\), to be at equilibrium, the temperature will be (a) \(1250 \mathrm{~K}\) (b) \(500 \mathrm{~K}\) (c) \(750 \mathrm{~K}\) (d) \(1000 \mathrm{~K}\)

If a gas at constant temperature and pressure expands, then its (a) internal energy decreases (b) entropy increases and then decreases (c) internal energy increases (d) internal energy remains constant

The enthalpy changes for the following processes are listed below. \(\mathrm{Cl}_{2}(\mathrm{~g})=2 \mathrm{C} 1(\mathrm{~g}) ; 242.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\mathrm{I}_{2}(\mathrm{~g})=21(\mathrm{~g}) ; 151.0 \mathrm{kJmol}^{-1}\) \(\mathrm{ICl}(\mathrm{g})=\mathrm{I}(\mathrm{g})+\mathrm{Cl}(\mathrm{g}) ; 211.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\mathrm{I}_{2}(\mathrm{~s})=\mathrm{I}_{2}(\mathrm{~g}) ; 62.76 \mathrm{~kJ} \mathrm{~mol}^{-1}\) Given that the standard states for iodine and chlorine are \(\mathrm{I}_{2}\) (s) and \(\mathrm{Cl},(\mathrm{g})\), the standard enthalpy of formation for \(\mathrm{ICl}(\mathrm{g})\) is [2006] (a) \(-14.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(-16.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(+16.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(+244.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Molar heat capacity at constant \(\mathrm{P}\) for a substance is equal to (a) \(\left(\delta_{U} / \delta_{\mathrm{T}}\right)_{\mathrm{v}}\) (b) \(\left(\delta_{H} / \delta_{T}\right)_{v}\) (c) \(\left(\delta_{U} / \delta_{\mathrm{T}}\right)_{\mathrm{p}}\) (d) \(\left(\delta_{H} / \delta_{\mathrm{T}}\right)_{\mathrm{p}}\)

Which of the following conditions are favourable for the feasibility of a reaction? (a) \(\Delta \mathrm{H}=-\mathrm{ve}, \mathrm{T} \Delta \mathrm{S}=+\mathrm{ve}\) (b) \(\Delta \mathrm{H}=-\mathrm{ve}, \mathrm{T} \Delta \mathrm{S}=-\mathrm{ve}, \mathrm{T} \Delta \mathrm{S}<\Delta \mathrm{H}\) (c) \(\Delta \mathrm{H}=+\mathrm{ve}, \mathrm{T} \Delta \mathrm{S}=+\mathrm{ve}, \mathrm{T} \Delta \mathrm{S}<\Delta \mathrm{H}\) (d) \(\Delta \mathrm{H}=+\mathrm{ve}, \mathrm{T} \Delta \mathrm{S}=+\mathrm{ve}, \mathrm{T} \Delta \mathrm{S}>\Delta \mathrm{H}\)

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