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A 72.8 L constant-volume cylinder containing \(7.41 \mathrm{g}\) He is heated until the pressure reaches 3.50 atm. What is the final temperature in degrees Celsius?

Short Answer

Expert verified
The final temperature of the gas is \(\approx 1977.17\) degrees Celsius.

Step by step solution

01

Convert mass of Helium to moles

To apply the Ideal Gas Law, the number of moles of the gas is needed. Helium (He) has a molar mass of \(4.0026 \mathrm{g / mol}\) approximately, so use the given mass \(7.41 \mathrm{g}\) to calculate the number of moles \(n\) using the formula: \(n = \frac{mass}{molar~mass}\) .
02

Convert pressure to the standard unit

Convert the given pressure from atmospheres to Pascal which is the standard unit of pressure for the Ideal Gas Law. The conversion factor is \(1 \text{ atm} = 1.01325 \times 10^5 \text{ Pa}\). So multiply the given pressure by the conversion factor.
03

Apply the Ideal Gas Law

Substitute the values of pressure \(P\), volume \(V\), moles \(n\) and the gas constant \(R = 8.314 \text{ J/mol K}\) into the Ideal Gas law and solve for the final temperature \(T\). Remember the temperature will be in Kelvin.
04

Convert final temperature to Celsius

Convert the final temperature in Kelvin to degrees Celsius using the formula: \(T_{\text{Celsius}} = T_{\text{Kelvin}} - 273.15\)

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Key Concepts

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

Moles Calculation
The Ideal Gas Law is a crucial part of understanding gas behaviors, and it often requires calculating the number of moles. This can be done using the mass of the gas and its molar mass. In this regard, Helium is quite straightforward as it has a molar mass of approximately 4.0026 g/mol.
If you are given a mass of helium gas, say 7.41 g, you can find the number of moles using the formula: \( n = \frac{\text{mass}}{\text{molar mass}} \).
Plug in the values: \( n = \frac{7.41 \text{ g}}{4.0026 \text{ g/mol}} \)
Doing this math will give you the number of moles of helium in the cylinder, which is a critical component for further calculations using the Ideal Gas Law.
Pressure Conversion
Pressure is another essential factor in the Ideal Gas Law. Atmospheric pressure units such as atmospheres (atm) are not the standard SI units used in calculations. The standard unit for pressure in the Ideal Gas Law is Pascal (Pa). Therefore, a conversion is necessary.
The conversion factor between these units is \(1 \text{ atm} = 1.01325 \times 10^5 \text{ Pa} \).
This means that if a problem gives you pressure in atm, you multiply it by this conversion factor to convert it to Pa.
  • This conversion ensures you are using the correct units for all terms in the Ideal Gas Law equation.
  • Given a pressure of 3.50 atm, convert it to Pa by multiplying: \(3.50 \times 1.01325 \times 10^5 \text{ Pa/atm} \).
This step is vital for accurate application of the Ideal Gas Law.
Temperature Conversion
Temperature in the Ideal Gas Law must be in Kelvin to ensure that calculations are accurate. While some problems might present temperature in Celsius, the conversion to Kelvin is a straightforward addition.
The conversion formula is: \( T_{\text{Kelvin}} = T_{\text{Celsius}} + 273.15 \).
The Kelvin scale starts at absolute zero, making it suitable for calculations in physics, especially gases.
  • After solving for the temperature in Kelvin using the Ideal Gas Law, you might need to convert it back to Celsius, particularly if the question specifies Celsius.
  • Convert Kelvin to Celsius using: \( T_{\text{Celsius}} = T_{\text{Kelvin}} - 273.15 \).
This conversion helps to bring back the temperature readings into a more familiar form when reporting final results.

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Most popular questions from this chapter

Calculate the total kinetic energy, in joules, of \(155 \mathrm{g} \mathrm{N}_{2}(\mathrm{g})\) at \(25^{\circ} \mathrm{C}\) and 1.00 atm. \([\text {Hint}:\) First calculate the average kinetic energy, \(\bar{e}_{k}\).

The Haber process is the principal method for fixing nitrogen (converting \(\mathrm{N}_{2}\) to nitrogen compounds). $$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g})$$ Assume that the reactant gases are completely converted to \(\mathrm{NH}_{3}(\mathrm{g})\) and that the gases behave ideally. (a) What volume of \(\mathrm{NH}_{3}(\mathrm{g})\) can be produced from 152 \(\mathrm{L} \mathrm{N}_{2}(\mathrm{g})\) and \(313 \mathrm{L}\) of \(\mathrm{H}_{2}(\mathrm{g})\) if the gases are measured at \(315^{\circ} \mathrm{C}\) and 5.25 atm? (b) What volume of \(\mathrm{NH}_{3}(\mathrm{g}),\) measured at \(25^{\circ} \mathrm{C}\) and\(727 \mathrm{mmHg},\) can be produced from \(152 \mathrm{L} \mathrm{N}_{2}(\mathrm{g})\) and \(313 \mathrm{L} \mathrm{H}_{2}(\mathrm{g}),\) measured at \(315^{\circ} \mathrm{C}\) and \(5.25 \mathrm{atm} ?\)

A sounding balloon is a rubber bag filled with \(\mathrm{H}_{2}(\mathrm{g})\) and carrying a set of instruments (the payload). Because this combination of bag, gas, and payload has a smaller mass than a corresponding volume of air, the balloon rises. As the balloon rises, it expands. From the table below, estimate the maximum height to which a spherical balloon can rise given the mass of balloon, \(1200 \mathrm{g} ;\) payload, \(1700 \mathrm{g}\) : quantity of \(\mathrm{H}_{2}(\mathrm{g})\) in balloon, \(120 \mathrm{ft}^{3}\) at \(0.00^{\circ} \mathrm{C}\) and \(1.00 \mathrm{atm}\); diameter of balloon at maximum height, 25 ft. Air pressure and temperature as functions of altitude are: $$\begin{array}{ccl} \hline \text { Altitude, km } & \text { Pressure, mb } & \text { Temperature, } \mathrm{K} \\ \hline 0 & 1.0 \times 10^{3} & 288 \\ 5 & 5.4 \times 10^{2} & 256 \\ 10 & 2.7 \times 10^{2} & 223 \\ 20 & 5.5 \times 10^{1} & 217 \\ 30 & 1.2 \times 10^{1} & 230 \\ 40 & 2.9 \times 10^{0} & 250 \\ 50 & 8.1 \times 10^{-1} & 250 \\ 60 & 2.3 \times 10^{-1} & 256 \\ \hline \end{array}$$

A particular gaseous hydrocarbon that is \(82.7 \%\) C and \(17.3 \%\) H by mass has a density of \(2.33 \mathrm{g} / \mathrm{L}\) at \(23^{\circ} \mathrm{C}\) and \(746 \mathrm{mm} \mathrm{Hg} .\) What is the molecular formula of this hydrocarbon?

Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2},\) is used to disinfect contact lenses. How many milliliters of \(\mathrm{O}_{2}(\mathrm{g})\) at \(22^{\circ} \mathrm{C}\) and \(752 \mathrm{mmHg}\) can be liberated from \(10.0 \mathrm{mL}\) of an aqueous solution containing \(3.00 \% \mathrm{H}_{2} \mathrm{O}_{2}\) by mass? The density of the aqueous solution of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is \(1.01 \mathrm{g} / \mathrm{mL}\) $$2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(1)+\mathrm{O}_{2}(\mathrm{g})$$

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