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One mole of monatomic ideal gas at \(\mathrm{T}(\mathrm{K})\) is expanded from \(1 \mathrm{~L}\) to \(2 \mathrm{~L}\) adiabatically under a constant external pressure of 1 atm the final temperature of the gas in Kelvin is (a) \(\mathrm{T}\) (b) \(\frac{\mathrm{T}}{2^{5 / 3-2}}\) (c) \(\mathrm{T}-\frac{2}{3 \times 0.0821}\) (d) \(T+\frac{3}{2 \times 0.0821}\)

Short Answer

Expert verified
The final temperature is \( \frac{T}{2^{2/3}} \). Choose option (b).

Step by step solution

01

Understanding the Adiabatic Process

An adiabatic process is one where no heat is exchanged with the surroundings. In the case of an ideal gas, the process is often described by the equation \( PV^\gamma = \text{constant} \), where \( \gamma \) is the heat capacity ratio \( C_p / C_v \). For a monatomic ideal gas, \( \gamma = \frac{5}{3} \).
02

Relating Pressure and Volume with Temperature

The adiabatic condition can also be expressed in terms of temperature and volume: \( TV^{\gamma-1} = \text{constant} \). In our case, since \( \gamma = \frac{5}{3} \), we can write \( TV^{2/3} = \text{constant} \).
03

Applying Initial and Final Conditions

Initially, we have \( TV_1^{2/3} \) and finally \( T'V_2^{2/3} \), where \( V_1 = 1 \) L and \( V_2 = 2 \) L. Hence: \( T \times 1^{2/3} = T' \times 2^{2/3} \). This simplifies to \( T = T' \times 2^{2/3} \).
04

Solving for Final Temperature \( T' \)

Rearrange the equation \( T = T' \times 2^{2/3} \) to solve for \( T' \): \[ T' = \frac{T}{2^{2/3}} \]. This form of the solution matches option (b) \( \frac{T}{2^{5/3-2}} \), since \( \frac{5}{3} - 2 = -\frac{1}{3} \).

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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 to simplify the study of gases. It is a hypothetical gas composed of many randomly moving point particles that interact only through elastic collisions. The concept is useful because it allows us to apply equations and models much more easily than with real gases.
Some key properties of an ideal gas include:
  • No volume: The particles in an ideal gas are considered point particles, meaning they themselves take up no space.
  • No interactions: The particles do not exert forces on one another except during elastic collisions.
  • Kinetic energy: The energy of the gas is entirely due to the motion of its particles, and temperature is a measure of this kinetic energy.
Let's use the ideal gas equation for many problems in chemistry and physics: \[ PV = nRT \]This equation relates pressure \( P \), volume \( V \), temperature \( T \), and number of moles \( n \) of an ideal gas, where \( R \) is the gas constant.
Monatomic Gas
A monatomic gas is an ideal gas consisting of single atoms, with no bonds between atoms as found in molecular gases. Common examples include noble gases like helium, neon, and argon. These gases are simple in structure, and their properties can be straightforwardly analyzed using the ideal gas law and kinetic theory.
Monatomic gases have some unique characteristics:
  • Heat capacity: Monatomic gases have constant heat capacities due to their simple structure. The molar heat capacity at constant volume \( C_v \) is \( \frac{3}{2}R \) for monatomic gases, where \( R \) is the gas constant.
  • Degrees of freedom: Because monatomic gases consist only of single atoms, they have fewer degrees of freedom compared to more complex molecules. Each atom moves in three dimensions (translational degrees of freedom), which accounts for their heat capacities.
The simplicity of monatomic gases makes them ideal candidates for studying thermodynamic processes in theoretical physics and chemistry.
Adiabatic Expansion
Adiabatic expansion is a process where a gas expands without exchanging heat with its surroundings. This means that all changes in the internal energy of the gas result from work done on or by the gas itself. Adiabatic processes are commonplace in thermodynamics and are characterized by the absence of heat transfer.
In mathematical terms, an adiabatic process is governed by the equation:\[ PV^\gamma = \text{constant} \]Here, \( P \) is pressure, \( V \) is volume, and \( \gamma \) is the heat capacity ratio. In adiabatic expansion:
  • The work is done by the gas, causing it to cool down.
  • Volume increases, and pressure decreases.
  • An ideal monatomic gas has \( \gamma = \frac{5}{3} \), allowing us to calculate final temperatures or pressures after expansion.
This concept reveals how temperature drops as volume increases due to work done by the gas in the absence of heat transfer.
Heat Capacity Ratio
The heat capacity ratio, denoted as \( \gamma \), is a crucial parameter in understanding thermodynamic processes like adiabatic expansion. It is the ratio of the molar heat capacity at constant pressure \( C_p \) to that at constant volume \( C_v \). For monatomic ideal gases, this is represented as:\[ \gamma = \frac{C_p}{C_v} = \frac{5}{3} \]This ratio is important because it determines how a gas behaves during processes where temperature, pressure, and volume change:
  • High \( \gamma \) values indicate that the gas will experience larger temperature changes during expansion or compression.
  • It influences the speed of sound in a gas, with higher values leading to a faster speed.
  • This ratio is crucial in calculating relations in adiabatic processes, like \( PV^\gamma = \text{constant} \).
Understanding the heat capacity ratio helps us predict how gases respond to changes in their environment, especially in engine cycles and atmospheric phenomena.

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Most popular questions from this chapter

Which of the following relation is/are incorrect? (a) \(\Delta \mathrm{G}=\Delta \mathrm{H}+\Delta \mathrm{nRT}\) (b) \(\Delta \mathrm{G}=\Delta \mathrm{H}+\mathrm{T} \Delta \mathrm{S}\) (c) \(\Delta \mathrm{G}=\Delta \mathrm{H}+\mathrm{T}[\delta\\{\Delta \mathrm{G}\\} / \delta \mathrm{T}]_{\mathrm{P}}\) (d) \(\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}\)

The standard enthalpy of combustion at \(25^{\circ} \mathrm{C}\) of \(\mathrm{H}_{2}\), \(\mathrm{C}_{6} \mathrm{H}_{10}\) and cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) are \(-241,-3800\) and \(-3920 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. Calculate heat of hydrogenation of cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{10}\right)\). (a) \(-161 \mathrm{kJmol}^{-1}\) (b) \(-131 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-121 \mathrm{kJmol}^{-1}\) (d) none

The enthalpy change of a reaction does not depend on (a) initial and final enthalpy change of reaction (b) state of reactants and products (c) different intermediate reactions (d) nature of reactants and products

For the system at equilibrium which of the following are correct? (a) On increasing the temperature of an endothermic reaction, the equilibrium shifts in forward direction because Q decreases. (b) On increasing the temperature of an endothermic reaction, the concentration in moles per litre of the reactants increases. (c) \(\log \mathrm{K}=\frac{1}{2.303 \mathrm{R}}\left(\Delta \mathrm{S}^{\circ}-\frac{\Delta \mathrm{H}^{\circ}}{\mathrm{T}}\right)\) (d) On increasing the temperature of an endothermic reaction, the equilibrium shifts in forward direction because \(\mathrm{K}\) increases.

For which of the following reactions, is \(\Delta \mathrm{H}\) equal to \(\Delta \mathrm{E} ?\) (a) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HI}(\mathrm{g})\) (b) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightarrow \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (c) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) (d) \(\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})\)

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