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Perfectly rigid containers each hold \(n\) moles of ideal gas, one being hydrogen \(\left(\mathrm{H}_{2}\right)\) and the other being neon \((\mathrm{Ne}) .\) If it takes \(300 \mathrm{~J}\) of heat to increase the temperature of the hydrogen by \(2.50^{\circ} \mathrm{C}\), by how many degrees will the same amount of heat raise the temperature of the neon?

Short Answer

Expert verified
The same amount of heat will raise the temperature of the neon by approximately \(4.16^{°}C\)

Step by step solution

01

Calculate the Molar Specific Heat of Hydrogen gas.

According to the given, heat (q) = 300J, change in temperature \( \Delta T = 2.5^{°}C = 2.5K \) (because 1 degree Celsius change = 1 K change), and \( n = 1 \) mole (for simplicity we consider n=1). Now, using formula q = nCΔT, we get \( C_{H_{2}} = q / n\Delta T = 300J / 1 * 2.5K = 120 \) J/mol.K
02

Calculate the ideal gas constant R using the Molar Specific heat.

Molar Specific Heat for any diatomic gas at constant volume \(C_v\) is \(5/2 R\). So, we have \(120 = 5/2 * R\). Let's solve this equation to find the value of R, we get \( R = 120 * 2 / 5 = 48 \) J/mol.K
03

Calculate the Molar Specific Heat of Neon gas.

Molar Specific Heat for any monoatomic gas at constant volume (\(C_v\)) is \(3/2 R\). Here R = 48 J/mol.K. Therefore, the molar specific heat of neon (\(C_{Ne}\)) = \(3/2 \)*48 = \(72 \) J/mol.K.
04

Calculate the rise in temperature of neon gas.

Now, we know that \(C_{Ne} = 72 \)J/mol.K. And using the same formula, \(q = nC\Delta T\), we can rearrange it to find \(\Delta T = q / nC = 300 / 1 * 72 = 4.16 K\).

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Key Concepts

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

Understanding the Ideal Gas Law
The ideal gas law is a fundamental principle that relates the pressure, volume, temperature, and amount of gas within a system. It's expressed by the equation:
\[ PV = nRT \]
where P stands for pressure, V for volume, n is the number of moles, R is the universal gas constant, and T represents the temperature in Kelvin. This law assumes that the gas particles are in constant, random motion and do not interact with each other, which is an approximation of real gases under certain conditions.
When solving problems, such as finding how much the temperature of a gas will increase as a result of heat transfer, the ideal gas law is often utilized to obtain other vital properties of the gas. In the case of the hydrogen and neon gases' exercise, knowing the number of moles and the specific heat capacity is essential for calculating the change in temperature when a certain amount of heat is added.
Exploring Molar Specific Heat
Molar specific heat, often denoted as C, is the amount of heat required to raise the temperature of one mole of a substance by one degree Kelvin. For gases, molar specific heat is important in thermodynamics as it can vary depending on whether the gas is monoatomic, diatomic, or polyatomic, and whether the process occurs at constant volume (Cv) or constant pressure (Cp).

The Specific Heat of Diatomic and Monoatomic Gases

In our exercise, hydrogen (a diatomic gas) and neon (a monoatomic gas) have different specific heats due to their atomic structures. Using the molar specific heat in conjunction with the ideal gas law allows us to calculate the temperature change when a gas absorbs or releases energy. In problems like the one given, understanding the relationship between heat transfer, molar specific heat, and temperature change (\[ q = nC\triangle T \]) is crucial in determining the desired temperature change for a known quantity of heat.
Heat Transfer in Gases
Heat transfer is the process of heat energy moving from one place to another and can occur through conduction, convection, or radiation. In the context of gases in a thermodynamic system, we mainly focus on the concept of heat added or removed from the system, which translates into a change in temperature.
In this exercise, the heat transfer to the gases is strictly conducted at a constant volume within rigid containers. Therefore, when heat is added, it directly affects the internal energy of the gas, which then translates to an elevation in temperature, assuming no work is done by the gas since the volume is held constant.
Using the specific heat capacity for gases at constant volume and the heat transfer equation (\[ q = nC\triangle T \]), we can solve for the change in temperature (\[ \triangle T \]) when a specif amount of heat (\[ q \]) is added. This overarching concept allows for the comparison of different gases' responses to heat transfer, as seen with hydrogen and neon.

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Most popular questions from this chapter

A cylindrical diving bell has a radius of \(750 \mathrm{~cm}\) and a height of \(2.50 \mathrm{~m}\). The bell includes a top compartment that holds an undersea adventurer. A bottom compartment separated from the top by a sturdy grating holds a tank of compressed air with a valve to release air into the bell, a second valve that can release air from the bell into the sea, a third valve that regulates the entry of seawater for ballast, a pump that removes the ballast to increase buoyancy, and an electric heater that maintains a constant temperature of \(20.0^{\circ} \mathrm{C}\). The total mass of the bell and all of its apparatuses is \(4350 \mathrm{~kg}\). The density of seawater is \(1025 \mathrm{~kg} / \mathrm{m}^{3}\). (a) An \(80.0 \mathrm{~kg}\) adventurer enters the bell. How many liters of seawater should be moved into the bell so that it is neutrally buoyant? (b) By carefully regulating ballast, the bell is made to descend into the sea at a rate of \(1.0 \mathrm{~m} / \mathrm{s}\). Compressed air is released from the tank to raise the pressure in the bell to match the pressure of the seawater outside the bell. As the bell descends, at what rate should air be released through the first valve? (Hint: Derive an expression for the number of moles of air in the bell \(n\) as a function of depth \(y ;\) then differentiate this to obtain \(d n / d t\) as a function of \(d y / d t .)\) (c) If the compressed air tank is a fully loaded, specially designed, \(600 \mathrm{ft}^{3}\) tank, which means it contains that volume of air at standard temperature and pressure ( \(0^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) ), how deep can the bell descend?

Estimate the number of atoms in the body of a \(50 \mathrm{~kg}\) physics student. Note that the human body is mostly water, which has molar mass \(18.0 \mathrm{~g} / \mathrm{mol}\), and that each water molecule contains three atoms.

Modern vacuum pumps make it easy to attain pressures of the order of \(10^{-13}\) atm in the laboratory. Consider a volume of air and treat the air as an ideal gas. (a) At a pressure of \(9.00 \times 10^{-14}\) atm and an ordinary temperature of \(300.0 \mathrm{~K}\), how many molecules are present in a volume of \(1.00 \mathrm{~cm}^{3} ?\) (b) How many molecules would be present at the same temperature but at 1.00 atm instead?

The vapor pressure is the pressure of the vapor phase of a substance when it is in equilibrium with the solid or liquid phase of the substance. The relative humidity is the partial pressure of water vapor in the air divided by the vapor pressure of water at that same temperature, expressed as a percentage. The air is saturated when the humidity is \(100 \%\). (a) The vapor pressure of water at \(20.0^{\circ} \mathrm{C}\) is \(2.34 \times 10^{3} \mathrm{~Pa}\). If the air temperature is \(20.0^{\circ} \mathrm{C}\) and the relative humidity is \(60 \%,\) what is the partial pressure of water vapor in the atmosphere (that is, the pressure due to water vapor alone)? (b) Under the conditions of part (a), what is the mass of water in \(1.00 \mathrm{~m}^{3}\) of air? (The molar mass of water is \(18.0 \mathrm{~g} / \mathrm{mol}\). Assume that water vapor can be treated as an ideal gas.)

Helium gas with a volume of \(3.20 \mathrm{~L}\), under a pressure of 0.180 atm and at \(41.0^{\circ} \mathrm{C}\), is warmed until both pressure and volume are doubled. (a) What is the final temperature? (b) How many grams of helium are there? The molar mass of helium is \(4.00 \mathrm{~g} / \mathrm{mol}\).

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