Problem 91
Question
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) \(225 \mathrm{~K}\)
Step-by-Step Solution
Verified Answer
The final temperature of the gas is 150 K.
1Step 1: Convert Initial Temperature to Kelvin
Initial temperature in Celsius is given as \(27^{\circ}C\). Convert it to Kelvin using the formula \(T(K) = T(^{\circ}C) + 273.15\). Thus, \(T_1 = 27 + 273.15 = 300.15\, K\).
2Step 2: Use Adiabatic Process Equation
For an adiabatic process, the equation \(W = C_v \cdot (T_1 - T_2)\) can be used, where \(W\) is the work done, \(C_v\) is the molar heat capacity at constant volume, and \(T_1, T_2\) are initial and final temperatures respectively.
3Step 3: Calculate the Change in Temperature
Given that \(W = 3\, kJ = 3000\, J\) and \(C_v = 20\, J/K/mol\), substitute values into the equation: \(3000 = 20 \cdot (300.15 - T_2)\).
4Step 4: Solve for Final Temperature
Rearrange the equation to find \(T_2\): \(3000 = 6003 - 20 \cdot T_2\). \(20 \cdot T_2 = 6003 - 3000\), hence \(20 \cdot T_2 = 3003\). Divide both sides by 20 to find \(T_2\): \(T_2 = \frac{3003}{20} = 150.15\, K\).
5Step 5: Choose the Correct Answer
Round \(T_2\) to the nearest integer. \(T_2 \approx 150\, K\). Therefore, the correct choice is (a) \(150\, K\).
Key Concepts
Adiabatic ProcessReversible ExpansionMolar Heat Capacity at Constant Volume
Adiabatic Process
In thermodynamics, an adiabatic process is one in which no heat is transferred into or out of the system. This means that the entire energy balance is due to work done by or on the system. The system's thermal energy changes without any exchange with its surroundings, making it a key concept in understanding energy conservation in isolated systems.
Adiabatic processes are characterized by the following features:
Adiabatic processes are characterized by the following features:
- There is no heat transfer, so the heat exchange, denoted by \(Q\), is zero.
- The energy change of the system is purely due to work done, which follows the first law of thermodynamics: \(\Delta U = W\), where \(\Delta U\) is the change in internal energy and \(W\) is the work done.
- In an adiabatic process with an ideal gas, the relation between pressure and volume (and consequently temperature changes) is more complex than in isothermal or isobaric processes.
Reversible Expansion
Reversible expansion is a theoretical concept that describes how systems undergo changes to their states. It is an ideal process, meaning it assumes the system changes in an infinitely slow manner, allowing it to remain in thermodynamic equilibrium. This concept is essential in understanding maximum potential work extraction.
- In a reversible process, changes are carried out so gradually that the system maintains equilibrium at every step.
- The energy transformations are perfectly efficient, allowing the system to do the maximum possible work on the surroundings.
- Mathematically, the work done by a gas during a reversible expansion can be described by the integral \(W = \int_{V_1}^{V_2} P \, dV\), where \(P\) is pressure and \(V\) is volume.
Molar Heat Capacity at Constant Volume
The molar heat capacity at constant volume, denoted as \(C_v\), is an essential concept in thermodynamics describing the heat required to raise the temperature of one mole of a substance by one degree Celsius (or Kelvin), without changing its volume. This form of heat capacity is relevant for processes fixed at constant volume, where no expansion work is done.
- \(C_v\) is useful for calculations involving gases, particularly ideal gases, where volume conditions are controlled.
- It is a direct measure of the energy needed to increase a mole of a gas's internal energy under volume constraints.
- The relationship between heat capacity, change in temperature, and energy change is given by \(\Delta Q = n C_v \Delta T\), where \(n\) is the number of moles.
Other exercises in this chapter
Problem 88
In monoatomic gases, ratio of specific heat at constant pressure to that at constant volume is (a) \(3 / 5\) (b) \(5 / 3\) (c) \(7 / 5\) (d) \(4 / 5\)
View solution Problem 89
The standard entropies of \(\mathrm{H}_{2}(\mathrm{~g}), \mathrm{I}_{2}(\mathrm{~s})\) and \(\mathrm{HI}(\mathrm{g})\) are \(130.6,116.7\) and \(206.3 \mathrm{~
View solution Problem 92
The change in entropy, in the conversion of one mole of water at \(373 \mathrm{~K}\) to vapour at the same temperature is (Latent heat of vaporization of water
View solution Problem 93
The direct conversion of \(\mathrm{A}\) to \(\mathrm{B}\) is difficult, hence it is carried out by the following path: Given \(\Delta \mathrm{S}(\mathrm{A} \lon
View solution