/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 66 One mole of a dilute diatomic ga... [FREE SOLUTION] | 91Ó°ÊÓ

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One mole of a dilute diatomic gas occupying a volume of 10.00 L expands against a constant pressure of 2.000 atm when it is slowly heated. If the temperature of the gas rises by \(10.00 \mathrm{K}\) and \(400.0 \mathrm{J}\) of heat are added in the process, what is its final volume?

Short Answer

Expert verified
The final volume of the diatomic gas after heating is approximately \(10.41 \mathrm{L}\).

Step by step solution

01

Identify the given information

We are given the following: - Initial volume (V1) = 10.00 L - Pressure (P) = 2.000 atm - Temperature change (ΔT) = 10.00 K - Heat added (Q) = 400.0 J - 1 mole of a dilute diatomic gas
02

Calculate the initial temperature

We can use the ideal gas law (PV = nRT) to find the initial temperature. Given that n = 1 mole, R (gas constant) = 0.0821 L * atm / (K * mol), we can rearrange the formula to find T1: T1 = PV / (nR) = (2.000 atm * 10.00 L) / (1 mole * 0.0821 L * atm / (K * mol)) ≈ 243.87 K
03

Calculate the final temperature

We know the temperature change (ΔT) is 10.00 K, so we can now find the final temperature (T2): T2 = T1 + ΔT = 243.87 K + 10.00 K ≈ 253.87 K
04

Use the combined gas law to find the final volume

The combined gas law formula is: (P1 * V1) / T1 = (P2 * V2) / T2 Since the pressure is constant (P1 = P2), we can simplify the formula to: V1 / T1 = V2 / T2 Now rearrange the formula to calculate the final volume (V2): V2 = (V1 * T2) / T1 = (10.00 L * 253.87 K) / 243.87 K ≈ 10.41 L
05

Conclusion

The final volume of the diatomic gas after heating is approximately 10.41 L.

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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 thermodynamics and chemistry, used to describe the state of an ideal gas. It is expressed as \(PV = nRT\), where:
  • \(P\) is the pressure of the gas.
  • \(V\) is the volume.
  • \(n\) is the number of moles.
  • \(R\) is the ideal gas constant.
  • \(T\) is the temperature in Kelvin.
This equation helps us relate pressure, volume, and temperature of a gas under ideal conditions. When using the Ideal Gas Law, it is crucial to ensure that units are consistent. For calculations involving liters, atmospheres, and moles, the common gas constant is 0.0821 L·atm/(K·mol).

In the exercise, the Ideal Gas Law was utilized to find the initial temperature of the gas using the given volume, pressure, and quantity of the gas. Knowing the initial state is essential for further calculations like determining changes in volume or temperature, especially when heat is applied.
Heat Transfer
Heat transfer is a process where energy is exchanged between a system and its surroundings due to a temperature difference. In the context of gases, heat can cause expansion or compression, affecting pressure and volume.
There are three modes of heat transfer:
  • Conduction: Direct transfer of heat through a material without movement of the material itself.
  • Convection: Transfer of heat by circulating fluid (liquid or gas) carrying heat with it.
  • Radiation: Transfer of energy by electromagnetic waves.
In the given problem, heat transfer occurs when 400.0 J of heat energy was added to the gas, causing it to expand. Understanding how this energy input affects the gas, in terms of internal energy and work done by or on the gas, allows us to analyze changes in its state, such as volume or temperature changes.
Gas Expansion
Gas expansion refers to an increase in the volume of a gas due to changes in temperature or pressure. When a gas is heated, the increased kinetic energy of its molecules causes them to occupy more space, resulting in expansion.
In the problem, the diatomic gas expands as heat is added, under constant pressure. This type of process is known as isobaric expansion. In such cases, the work done by the gas can be calculated by the formula \(W = P \Delta V\). However, since we are focusing on finding the final volume, the formula helpfully simplifies due to constant pressure, allowing us to utilize the relationship between volume and temperature provided by the combined gas law:

\[\frac{V_1}{T_1} = \frac{V_2}{T_2}\]where \(V_1\) and \(V_2\) are initial and final volumes, and \(T_1\) and \(T_2\) are initial and final temperatures, respectively. This highlights how temperature affects volume in an ideal gas under constant pressure.
Diatomic Gas
Diatomic gases are a type of gas composed of molecules made up of two atoms. Common examples include oxygen \((O_2)\) and nitrogen \((N_2)\). These gases have unique rotational and vibrational modes compared to monatomic gases, affecting their thermodynamic behavior.
The heat capacity of a diatomic gas can be slightly different due to these additional degrees of freedom. This means they require more energy to achieve the same change in temperature compared to monatomic gases. In the exercise, understanding that the gas is diatomic is crucial as it influences heat capacity and thus the energy required for temperature changes.
While not directly impacting the volume calculation here, knowing the type of gas helps grasp the broader thermodynamic principles at play. It also helps in understanding how energy input can affect different gases, ensuring correct application of thermodynamic laws in varied situations.

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Most popular questions from this chapter

On an adiabatic process of an ideal gas pressure, volume and temperature change such that \(p V^{\gamma}\) is constant with \(\gamma=5 / 3\) for monatomic gas such as helium and \(\gamma=7 / 5\) for diatomic gas such as hydrogen at room temperature. Use numerical values to plot two isotherms of 1 mol of helium gas using ideal gas law and two adiabatic processes mediating between them. Use T_{1}=500 \mathrm{K}, V_{1}=1 \mathrm{L}, \text { and } T_{2}=300 \mathrm{K} \text { for your plot. }

Is it possible for \(\gamma\) to be smaller than unity?

A metallic container of fixed volume of \(2.5 \times 10^{-3} \mathrm{m}^{3}\) immersed in a large tank of temperature \(27^{\circ} \mathrm{C}\) contains two compartments separated by a freely movable wall. Initially, the wall is kept in place by a stopper so that there are 0.02 mol of the nitrogen gas on one side and 0.03 mol of the oxygen gas on the other side, each occupying half the volume. When the stopper is removed, the wall moves and comes to a final position. The movement of the wall is controlled so that the wall moves in infinitesimal quasi-static steps. (a) Find the final volumes of the two sides assuming the ideal gas behavior for the two gases. (b) How much work does each gas do on the other? (c) What is the change in the internal energy of each gas? (d) Find the amount of heat that enters or leaves each gas.

A car tire contains \(0.0380 \mathrm{m}^{3}\) of air at a pressure of \(2.20 \times 10^{5} \mathrm{Pa}\) (about 32 psi). How much more internal energy does this gas have than the same volume has at zero gauge pressure (which is equivalent to normal atmospheric pressure)?

A cylinder containing three moles of a monatomic ideal gas is heated at a constant pressure of 2 atm. The temperature of the gas changes from \(300 \mathrm{K}\) to \(350 \mathrm{K}\) as a result of the expansion. Find work done (a) on the gas; and (b) by the gas.

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