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The internal energy change (in \(\mathrm{J}\) ) when \(90 \mathrm{~g}\) of water undergoes complete evaporation at \(100^{\circ} \mathrm{C}\) is \(.\) (Given : \(\Delta \mathrm{H}_{\text {vap }}\) for water at \(373 \mathrm{~K}=41 \mathrm{~kJ} / \mathrm{mol}, \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) )

Short Answer

Expert verified
The internal energy change is 40.5 kJ.

Step by step solution

01

Understand the Problem

The problem involves calculating the internal energy change during the evaporation of water at 100°C. We are given the enthalpy of vaporization and need to find the internal energy change. This can be done using the formula: \[ \Delta U = \Delta H - P\Delta V \]where \( P\Delta V = nR\Delta T \). However, \( \Delta T \) is zero because the process happens at a constant temperature, simplifying to \(P\Delta V = nRT\text{ (only for ideal gas approximation)} \).

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Key Concepts

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

Internal Energy Change
The change in internal energy, denoted as \( \Delta U \), reflects how much energy is absorbed or released within a system during a process, like the evaporation of water. During a phase change, such as when water transforms from liquid to vapor, energy is needed to overcome intermolecular forces. This energy input does not contribute to a temperature change but rather alters the state of the substance.
In the exercise, we calculate \( \Delta U \) using the formula:
  • \( \Delta U = \Delta H - P\Delta V \)
Here, \( \Delta H \) is the enthalpy change during vaporization, and \( P\Delta V \) accounts for the work done by the system as it expands, governed by the ideal gas approximation. Understanding internal energy change is key in thermodynamics, as it provides insight into the energy balance and behavior of substances under different conditions.
Enthalpy of Vaporization
Enthalpy of vaporization, \( \Delta H_{\text{vap}} \), is the energy required to convert a substance from a liquid to a gas at constant temperature and pressure. For water, this process occurs at 100°C or 373K, during which each mole of water absorbs heat.
  • Given that \( \Delta H_{\text{vap}} = 41 \text{ kJ/mol} \) for water, this value is intrinsic to water's ability to vaporize.
Enthalpy changes, like \( \Delta H_{\text{vap}} \), are significant because they reflect the strength of intermolecular forces; in water's case, strong hydrogen bonds. Evaluating \( \Delta H_{\text{vap}} \) allows us to quantify the energy needed for phase changes in chemical processes and infer the energy dynamics involved.
Ideal Gas Approximation
The ideal gas approximation simplifies calculations by assuming certain conditions under which a gas behaves perfectly. For gases, this means they're considered to have:
  • Elastic collisions
  • No intermolecular forces
  • Infinitesimal particle volume
For the problem at hand, the ideal gas law, \( PV = nRT \), aids in calculating work done by the system as water vaporizes, expressed in the term \( P\Delta V = nRT \). Given that \( \Delta T = 0 \) (as the temperature remains constant during vaporization), the term simplifies calculations without loss of accuracy.While real gases exhibit behaviors deviating from this model, at high temperatures and low pressures, the ideal gas approximation can effectively predict gas characteristics. This framework is integral to problem-solving in thermodynamics where approximations are necessary for practical calculations.

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Most popular questions from this chapter

The enthalpy of vapourization of liquid is \(30 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and entropy of vapourization is \(75 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}\). The boiling point of the liquid at \(1 \mathrm{~atm}\) is [2004S] (a) \(250 \mathrm{~K}\) (b) \(400 \mathrm{~K}\) (c) \(450 \mathrm{~K}\) (d) \(600 \mathrm{~K}\)

The polymerisation of ethylene to linear polyethylene is represented by the reaction [1994 - 2 Marks] \(n \mathrm{CH}_{2}=\mathrm{CH}_{2} \longrightarrow \mathrm{E} \mathrm{CH}_{2}-\mathrm{CH}_{2} \frac{1}{\pi}\) where \(n\) has a large integral value. Given that the average enthalpies of bond dissociation for \(\mathrm{C}=\mathrm{C}\) and \(\mathrm{C}-\mathrm{C}\) at \(298 \mathrm{~K}\) are \(+590\) and \(+331 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}\), respectively, calculate the enthalpy of polymerisation per mole of ethylene at \(298 \mathrm{~K}\).

The true statement amongst the following is : [Main Jan. 09, 2020 (II)] (a) Both \(\Delta \mathrm{S}\) and \(\mathrm{S}\) are functions of temperature. (b) Both \(\mathrm{S}\) and \(\Delta \mathrm{S}\) are not functions of temperature. (c) \(\mathrm{S}\) is not a function of temperature but \(\Delta \mathrm{S}\) is a function of temperature. (d) \(\mathrm{S}\) is a function of temperature but \(\Delta \mathrm{S}\) is not a function of temperature.

Two moles of a perfect gas undergo the following processes: (a) a reversible isobaric expansion from \((1.0 \mathrm{~atm}, 20.0 \mathrm{~L})\) to \((1.0 \mathrm{~atm}, 40.0\) L); (b) a reversible isochoric change of state from \((1.0 \mathrm{~atm}, 40.0 \mathrm{~L})\) to \((0.5\) atm, \(40.0 \mathrm{~L}\) ); (c) a reversible isothermal compression from \((0.5 \mathrm{~atm}, 40.0 \mathrm{~L})\) to \((1.0\) atm, \(20.0 \mathrm{~L}\) ). (i) Sketch with labels each of the processes on the same \(P-V\) diagram. (ii) Calculate the total work \((W)\) and the total heat change \((q)\) involved in the above processes. (iii) What will be the values of \(\Delta U, \Delta H\) and \(\Delta S\) for the overall process?

One mole of a non-ideal gas undergoes a change of state ( \(2.0\) atm, 3.0L, 95K) \(\rightarrow(4.0\) atm \(, 5.0 \mathrm{~L}, 245 \mathrm{~K})\) with a change in internal energy, \(\Delta U=30.0 \mathrm{~L} \mathrm{~atm} .\) The change in enthalpy \((\Delta H)\) of the process in \(\mathrm{L}\) atm is [2002S] (a) \(40.0\) (b) \(42.3\) (c) \(44.0\) (d) not defined, because pressure is not constant

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