91Ó°ÊÓ

Helium gas, a noble gas often used in balloons, exhibits ideal gas behavior under standard conditions. In this exercise, we examine helium gas that initially occupies a volume of \(2.60\, \text{L}\) under a pressure of \(1.30\, \text{atm}\) at a temperature of \(41.0^{\circ}\text{C}\). The problem entails calculating the new temperature when both the pressure and volume are doubled, as well as the mass of helium in the system.
First, we convert the given temperature from degrees Celsius to Kelvin because kelvin is the standard unit used in gas calculations. Helium's chemical property of low molar mass \(4.00\, \text{g/mol}\) makes it less dense than most other gases, facilitating its usage in buoyancy-related applications. By employing the Ideal Gas Law and subsequent calculations, we determine the changes in state variables of the helium gas.

The Combined Gas Law is a useful equation that links the initial and final states of a gas under varying conditions of pressure, volume, and temperature. This law is crucial for this exercise, where helium gas's pressure and volume are both doubled.
The Combined Gas Law formula is given as: \[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \] In this expression, \(P_1\), \(V_1\), and \(T_1\) are the initial pressure, volume, and temperature, respectively, while \(P_2\), \(V_2\), and \(T_2\) are their final states.
By solving for \(T_2\), the final temperature, we employ the values from the initial and altered states to establish the conditions under which the helium gas reaches equilibrium after the change in pressure and volume.

Calculating the number of moles and consequently the mass of a gas is fundamental in comprehending gas reactions and quantities. Using helium gas, the exercise illustrates calculating moles using the Ideal Gas Law equation \( PV = nRT \), where \( n \) represents moles.
Substituting the known values allows us to simplify: \[ n = \frac{P_1V_1}{RT_1} \]. This gives us the quantity of helium in terms of moles, which is \(0.132\, \text{mol}\).
Subsequent steps involve multiplying the moles of helium by its molar mass \(4.00\, \text{g/mol}\) to derive the mass in grams. These calculations help manifest the physical reality of chemical amounts into measurable quantities, crucial for preparing reactions, understanding yields, and more.

Converting temperatures to Kelvin is an essential process in many chemistry calculations, particularly those involving gases. Kelvin is the absolute temperature scale used in the study of gases because it allows for direct proportional relationships in equations.
The conversion from Celsius to Kelvin is straightforward: add 273.15 to the Celsius measurement. In this problem, the conversion takes place to transform \(41.0^{\circ}\text{C}\) into \(314.15\, \text{K}\).
This new unit becomes fundamental throughout the calculations, enabling seamless usage in equations like the Ideal Gas Law and the Combined Gas Law. Understanding and applying this conversion helps ensure accurate results in chemical computations and analysis, embodying a foundational step in virtually any thermodynamics problem involving gases.