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Chapter 11: Problem 114

If \(8.50\) moles of He gas occupy a volume of \(25.0 \mathrm{~L}\) at \(28^{\circ} \mathrm{C}\), what pressure in atmospheres does the gas exert?

Short Answer

Expert verified
The pressure exerted by the He gas is approximately 8.33 atmospheres. To calculate this, we used the Ideal Gas Law equation, converted the given temperature to Kelvin, and used the correct value of the ideal gas constant for pressure in atmospheres.

Step by step solution

01

Convert temperature to Kelvin

To convert the given temperature from Celsius to Kelvin, add 273.15. In this case, the temperature is given as \(28^{\circ}C\). \[T (K) = T (^\circ C) + 273.15\] \[T (K) = 28 + 273.15 = 301.15\, K\]
02

Determine the ideal gas constant

Since we are asked to find the pressure in atmospheres, we need to determine the appropriate value of the ideal gas constant, R, to use in the Ideal Gas Law equation. The ideal gas constant for pressure in atmospheres is: \[R = 0.0821\, \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}}\]
03

Use the Ideal Gas Law to find the pressure

Now we have all the values needed to solve for pressure using the Ideal Gas Law equation: \[PV = nRT\] Plug in the values: \[P \cdot 25.0\text{ L} = 8.50\text{ mol} \cdot 0.0821\, \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}} \cdot 301.15\, K\]
04

Isolate pressure and calculate the value

Next, we need to isolate the pressure variable by dividing both sides of the equation by the value of the volume: \[\frac{P \cdot 25.0\text{ L}}{25.0\text{ L}} = \frac{8.50\text{ mol} \cdot 0.0821\, \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}} \cdot 301.15\, K}{25.0\text{ L}}\] \[P = \frac{8.50 \cdot 0.0821 \cdot 301.15}{25.0}\, \text{atm}\] \[P = 8.3263\, \text{atm}\]
05

Report the pressure in the correct unit

Finally, the pressure of the He gas in atmospheres is: \[P \approx 8.33\, \text{atm}\] Thus, the pressure exerted by the gas is about 8.33 atmospheres.

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Most popular questions from this chapter

A balloon is filled with \(\mathrm{H}_{2}\) gas to a volume of \(1610.2 \mathrm{~mL}\). The pressure of the gas in the balloon is \(745.4 \mathrm{~mm} \mathrm{Hg}\), and the temperature is \(22.7^{\circ} \mathrm{C}\). What is the mass in grams of the \(\mathrm{H}_{2}\) in the balloon?

A balloon filled with He gas and another balloon filled with \(\mathrm{H}_{2}\) gas have the same values for \(P\) and \(\bar{T}\). (a) The density of the He gas is greater than the density of the \(\mathrm{H}_{2}\) gas. How can you prove this using the ideal gas law? (b) How much more dense than the \(\mathrm{H}_{2}\) gas is the He gas?

A sample of gas at \(25^{\circ} \mathrm{C}\) and \(759.0 \mathrm{~mm} \mathrm{Hg}\) has a volume of \(1.58 \mathrm{~L}\). If the temperature is raised to \(35^{\circ} \mathrm{C}\) but the pressure is held constant at \(759.0 \mathrm{~mm} \mathrm{Hg}\), will the volume increase or decrease? Explain your answer.

What is the density in grams per liter of nitrogen gas at STP?

A 1.56-g sample of gas is contained in a \(250.0-\mathrm{mL}\) cylinder. Its pressure is \(1255.6 \mathrm{~mm} \mathrm{Hg}\), and its temperature is \(22.7{ }^{\circ} \mathrm{C}\). (a) What is the molar mass of the gas? (b) Combustion analysis reveals the empirical formula of this gas to be \(\mathrm{NO}_{2}\). What is the molecular formula?
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