A 1.00 -mole sample of ammonia at 14.0 atm and \(25^{\circ} \mathrm{C}\) in a cylinder fitted with a movable piston \(\mathrm{ex}\) pands against a constant external pressure of 1.00 atm. At equilibrium, the pressure and volume of the gas are 1.00 atm and 23.5 L, respectively. (a) Calculate the final temperature of the sample. (b) Calculate \(q, w,\) and \(\Delta E\) for the process. The specific heat of ammonia is \(0.0258 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The final temperature of the ammonia sample is 286K. The values of heat exchange (q), work done (w), and the change in internal energy (ΔE) for the process are -0.309 kJ, -2.205 kJ, and -2.514 kJ, respectively.

Step by step solution

01

Calculation of final temperature

Using the Ideal Gas Law \(PV = nRT\), where \(P=1.00 atm\), \(V=23.5 L\), \(n=1.00 mol\), and \(R=0.0821 L·atm/K·mol\), let’s calculate the final temperature \(T_f\). \(T_f = \frac{PV}{nR} = \frac{(1.00 atm)(23.5 L)}{(1.00 mol)(0.0821 L·atm/K·mol)} = 286K\)
02

Calculating work done

The work done \(w\) by the system can be calculated using the formula \(w = – P_ex(V_f – V_i)\), where \(P_ex\) = external pressure = 1.00 atm, \(V_f\) = final volume = 23.5 L, and \(V_i\) = initial volume can be calculated with initial conditions by using the Ideal Gas Law, which gives \(V_i = \frac{nRT_i}{P_i} = \frac{(1.00 mol)(0.0821 L·atm/K·mol)(298 K)}{14.0 atm} = 1.752 L\). So, \(w = - (1.00 atm)(23.5 L - 1.752 L) = -21.748 L·atm = -2.205 kJ\) (as 1 L·atm = 101.3 J).
03

Calculating heat exchange

The heat exchange \(q\) can be calculated using the formula \(q = nC (T_f – T_i)\), where \(n=1.00 mol\), \(C=0.0258 kJ/mol·K\), \(T_f=286K\), and \(T_i = 25°C = 298K\). Therefore, \(q = (1.00 mol)(0.0258 kJ/mol·K)(286K - 298 K) = -0.309 kJ\).
04

Calculating change in internal energy

According to the First Law of Thermodynamics, the change in internal energy ΔE = q + w, so ΔE = -0.309 kJ – 2.205 kJ = -2.514 kJ.

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation in thermochemistry, encapsulating how pressure (P), volume (V), temperature (T), and the amount of substance (n) are interrelated in a gas that behaves ideally. The equation is given by PV = nRT, where R represents the ideal gas constant.

Considering the exercise problem, the Ideal Gas Law helps us determine the final temperature of the gas after expansion by rearranging the formula to solve for T. This principle is also used to find the initial volume of the gas before expansion, allowing us to grasp how the changing conditions affect the gas behavior and understanding that real gases approximate ideal behavior under many conditions.

In application, this law is essential for predicting the behavior of gases during chemical reactions or physical transformations, as it correlates the measurable properties of temperature, volume, and pressure with the amount of gas present.
First Law of Thermodynamics
The First Law of Thermodynamics, a variation of the Law of Conservation of Energy, states that energy cannot be created or destroyed in an isolated system. In the context of thermochemistry, it is usually expressed as ΔE = q + w, where ΔE is the change in internal energy, q is the heat exchanged, and w is the work done by the gas.

When working through the problem at hand, we utilize this principle to relate the work done by the gas during expansion and the heat absorbed or released to calculate the overall change in the gas’s internal energy. By recognizing that the work and heat correlate to changes in the gas's environment, we can better understand how energy is transformed and conserved during physical processes. This first law serves as the foundation for analyzing energy changes in chemical reactions and physical changes.
Heat Exchange in Gases
Heat exchange in gases refers to the transfer of energy due to a temperature difference. We use the formula q = nCΔT to calculate the heat exchanged (q), where n is the number of moles, C is the molar specific heat capacity at constant pressure or volume, and ΔT is the change in temperature.

In our exercise, we applied this formula to find the amount of heat absorbed or released by the gas. Since heat flow occurs to achieve thermal equilibrium, understanding this concept allows us to explain the energy exchange with the surroundings. Additionally, whether the gas absorbs or releases heat can give us insights into the nature of the process—endothermic or exothermic—and the impact on the system's overall energy.
Work Done by Gases
Work done by gases during expansion or compression is a product of the gas applying force over a distance. For gases expanding against a constant external pressure, the work (w) can be calculated by using the equation w = -Pex(Vf - Vi), wherein Pex stands for the external pressure, Vf is the final volume, and Vi is the initial volume.

In the given problem, we determined the work done by the expanding ammonia using this work formula. The work is negative because the system (gas) does work on the surroundings by expanding, reflecting energy leaving the system. This aspect of thermochemistry is critical in understanding how gases can perform mechanical work, which is a crucial concept in fields such as engineering and meteorology.

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