To minimize the rate of evaporation of the tungsten filament, \(1.4 \times 10^{-5}\) mol of argon is placed in a \(600-\mathrm{cm}^{3}\) light-bulb. What is the pressure of argon in the lightbulb at \(23^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The pressure of argon inside the lightbulb at \(23^\circ C\) is approximately \(5.38 \times 10^{-5}\) atm.

Step by step solution

01

Convert the temperature to Kelvin

To calculate the pressure of argon, we need the temperature in Kelvin. To convert Celsius to Kelvin, add 273.15 to the given temperature. \(T(K) = 23^\circ C + 273.15\) \(T(K) = 296.15 K\)
02

Convert volume to liters

The volume of the lightbulb is given in cubic centimeters (cm³), but we need the volume in liters (L) to use the Ideal Gas Law. To convert cm³ to L, divide the volume by 1000. \(V(L) = 600 \text{ cm}^3 \times \frac{1 \text{ L}}{1000 \text{ cm}^3}\) \(V(L) = 0.600 \text{ L}\)
03

Find the universal gas constant value

The universal gas constant (R) has various different units. We will use the unit that has L atm / mol K to be consistent with our given values. \(R = 0.0821 \frac{\text{L atm}}{\text{mol K}}\)
04

Apply the Ideal Gas Law formula

Now, we have all the information we need to apply the Ideal Gas Law formula and solve for pressure (P). \(PV = nRT\) Rearrange the formula to solve for P: \(P = \frac{nRT}{V}\) Plug in the values: \(P = \frac{(1.4 \times 10^{-5} \text{ mol})(0.0821 \frac{\text{L atm}}{\text{mol K}})(296.15 \text{ K})}{0.600 \text{ L}}\)
05

Calculate the pressure

Now, we simply perform the calculation in the formula. \(P = \frac{(1.4 \times 10^{-5})(0.0821)(296.15)}{0.600}\) \(P = 5.38 \times 10^{-5} \text{ atm}\) The pressure of argon inside the lightbulb at \(23^\circ C\) is approximately \(5.38 \times 10^{-5}\) atm.

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Most popular questions from this chapter

An ideal gas at a pressure of 1.50 atm is contained in a bulb of unknown volume. A stopcock is used to connect this bulb with a previously evacuated bulb that has a volume of 0.800 \(\mathrm{L}\) as shown here. When the stopcock is opened, the gas expands into the empty bulb. If the temperature is held constant during this process and the final pressure is 695 torr, what is the volume of the bulb that was originally filled with gas?

You have a gas at \(25^{\circ} \mathrm{C}\) confined to a cylinder with a movable piston. Which of the following actions would double the gas pressure? \((\mathbf{a})\) Lifting up on the piston to double the volume while keeping the temperature constant; \((\mathbf{b})\) Heating the gas so that its temperature rises from \(25^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) , while keeping the volume constant; \((\mathbf{c})\) Pushing down on the piston to halve the volume while keeping the temperature constant.

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Consider a mixture of two gases, \(A\) and \(B,\) confined in a closed vessel. A quantity of a third gas, \(C,\) is added to the same vessel at the same temperature. How does the addition of gas C affect the following: (a) the partial pressure of gas A, (b) the total pressure in the vessel, (c) the mole fraction of gas B?

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