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2 mole of an ideal gas at \(27^{\circ} \mathrm{C}\) temperature is expanded reversibly from \(2 \mathrm{~L}\) to \(20 \mathrm{~L}\). Find entropy change in cal. \((\mathrm{R}=2 \mathrm{cal} / \mathrm{mol} \mathrm{K})\) (a) \(92.1\) (b) 0 (c) 4 (d) \(9.2\)

Short Answer

Expert verified
(d) 9.2 cal.

Step by step solution

01

Convert Temperature to Kelvin

The given temperature is \(27^{\circ} \text{C}\). We convert it to Kelvin by adding 273.15. So, the temperature is \(27 + 273.15 = 300.15\, \text{K}\). For simplicity in calculations, we'll use \(300\, \text{K}\).
02

Recall the Formula for Entropy Change

The formula for the change in entropy \(\Delta S\) during a reversible isothermal process for an ideal gas is given by:\[\Delta S = n R \ln \left( \frac{V_f}{V_i} \right)\]where \(n\) is the number of moles, \(R\) is the ideal gas constant, \(V_f\) is the final volume, and \(V_i\) is the initial volume.
03

Plug Values into the Entropy Change Formula

Substitute the given values into the formula: \[n = 2, \, R = 2 \, \text{cal/mol K}, \, V_f = 20 \, \text{L}, \, V_i = 2 \, \text{L}\].So, the entropy change \(\Delta S\) is:\[\Delta S = 2 \times 2 \ln \left( \frac{20}{2} \right)\]
04

Calculate the Natural Logarithm

Calculate \(\ln \left( \frac{20}{2} \right) = \ln(10)\). The value of \(\ln(10)\) is approximately 2.302.
05

Calculate the Entropy Change

Using the natural logarithm calculated:\[\Delta S = 4 \times 2.302 = 9.208 \approx 9.2 \, \text{cal}\].
06

Choose the Closest Answer

From the available options, the closest answer to our calculated value (9.208) is option (d) 9.2.

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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 where the gas molecules do not interact with each other except through elastic collisions. This means there are no intermolecular forces except during these collisions.

Ideal gases are significant because they help us understand how real gases behave under common conditions. In many cases, real gases perform very similarly to an ideal gas, especially at high temperatures and low pressures. It’s easier to calculate and predict the properties of gases by assuming they are ideal when using conditions that are close enough to those where real gases deviate only slightly.
  • The molecules occupy negligible space when compared to the container they are in.
  • There are no forces of attraction between particles, aside from during collisions.
  • The particles are in random motion and collisions are completely elastic.
Using these assumptions, we apply the ideal gas law which is expressed as \( PV = nRT \). This formula is integral for various calculations, such as those involving changes in entropy that we encounter in many thermodynamic problems.
Reversible Process
In thermodynamics, a reversible process is a hypothetical situation where the system and surroundings can be returned to their initial states without any change in the universe. It is an idealization and is not practically achievable due to energy dissipation in real processes.

A reversible process is carried out in an infinitely slow manner to ensure the system remains in a state of equilibrium with its surroundings at all times. This allows each step of the process to be reversed by an infinitesimal change. Although real processes are irreversible, reversible processes set a theoretical limit to the efficiency and performance of thermodynamic systems.

Key characteristics:
  • Transition of system changes so slowly that the system can adjust to equilibrium conditions at each phase of the process.
  • No entropy is produced within the system, although entropy might be transferred from the system to the surroundings or vice versa.
  • A reversible process can be reverted without changing the overall entropy of the universe.
This concept is essential in deriving important thermodynamic equations and forms the theoretical underpinnings for determining maximum efficiency.
Isothermal Process
An isothermal process is a thermodynamic change wherein the temperature of a system remains constant throughout the process. This happens when heat exchange with the surroundings compensates exactly for the work done, keeping the internal energy unchanged.

Isothermal processes are key in understanding ideal gas behavior, particularly in entropy change calculations where temperature consistency is crucial. For an ideal gas, it is often connected to reversible processes allowing for straightforward mathematical manipulation and analysis.
  • Maintains constant temperature due to heat transfer balance.
  • Generally occurs slowly to allow the system to continuously exchange heat with surroundings to keep the temperature stable.
  • Commonly analyzed in the context of an ideal gas where internal energy remains unchanged.
This process is crucial for simplification when using the formula for entropy change: \[ \Delta S = nR \ln \left( \frac{V_f}{V_i} \right) \], where the consistency in temperature allows this equation to hold true.
Natural Logarithm
The natural logarithm, denoted as \( \ln \), is a logarithmic function with a base of \( e \), where \( e \) is an irrational constant approximately equal to 2.71828. In the context of thermodynamics, the natural logarithm is particularly useful when dealing with exponential relationships and decay patterns.

In many thermodynamic equations, you will see the natural logarithm used due to its presence in growth and decay problems, helping to simplify the mathematics involved. Especially when calculating entropy changes, the natural logarithm helps express the quantitative relationships between states in a more manageable form.

Notable points to remember:
  • The natural logarithm function is the inverse of the exponential function \( e^x \).
  • It is commonly used in calculations involving proportionate growth or decay.
  • In reversible isothermal processes for an ideal gas, it allows the expression and calculation of entropy change in simplified terms.
Using this, we calculate entropy changes effectively, as seen in the formula: \( \Delta S = nR \ln \left( \frac{V_f}{V_i} \right) \), demonstrating its utility as it was in our solution.

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Most popular questions from this chapter

Two moles of an ideal gas are compressed at \(300 \mathrm{~K}\) from a pressure of 1 atm to a pressure of 2 atm. The change in free energy is (a) \(5.46 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(2.46 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(3.46 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(8.46 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

In thermodynamics, a process is called reversible when (a) surroundings and system change into each other (b) there is no boundary between system and surroundings (c) the surroundings are always in equilibrium with the system (d) the system changes into the surroundings sponta neously

Considering entropy (S) as a thermodynamic parameter, the criterion for the spontaneity of any process is (a) \(\Delta \mathrm{S}_{\text {system }}+\Delta \mathrm{S}_{\text {surroundings }}>0\) (b) \(\Delta \mathrm{S}_{\text {system }}-\Delta \mathrm{S}_{\text {surroundings }}>0\) (c) \(\Delta \mathrm{S}_{\text {system }}>0\) (d) \(\Delta \mathrm{S}_{\text {surroundings }}>0\)

Which of the following conditions may lead to a nonspontaneous change? (a) \(\Delta \mathrm{H}=-\mathrm{ve} ; \Delta \mathrm{S}=+\mathrm{ve}\) (b) \(\Delta \mathrm{H}=-\mathrm{ve} ; \Delta \mathrm{S}=-\mathrm{ve}\) (c) \(\Delta \mathrm{H}\) and \(\Delta \mathrm{S}\) are both \(+\mathrm{ve}\) (d) \(\Delta H=+v e ; \Delta S=-v e\)

Calculate the enthalpy change for the combustion of cyclopropane at \(298 \mathrm{~K}\), if the enthalpy of formation \(\mathrm{CO}_{2}(\mathrm{~g}), \mathrm{H}_{2} \mathrm{O}(1)\) and propene \((\mathrm{g})\) are \(-393.5,-385.8\) and \(20.42 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. The enthalpy of isomerization of cyclopropane to propene is \(-33.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (a) \(1802 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(2091 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(2196 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) none

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