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Chapter 10: Problem 34

Calculate each of the following quantities for an ideal gas: (a) the volume of the gas, in liters, if \(1.50 \mathrm{~mol}\) has a pressure of $126.7 \mathrm{kPa}\( at a temperature of \)-6^{\circ} \mathrm{C} ;(\mathbf{b})$ the absolute temperature of the gas at which \(3.33 \times 10^{-3}\) mol occupies \(478 \mathrm{~mL}\) at \(99.99 \mathrm{kPa} ;(\mathbf{c})\) the pressure, in pascals, if \(0.00245 \mathrm{~mol}\) occupies \(413 \mathrm{~mL}\) at \(138^{\circ} \mathrm{C} ;(\mathbf{d})\) the quantity of gas, in moles, if 126.5 L at \(54^{\circ} \mathrm{C}\) has a pressure of \(11.25 \mathrm{kPa}\).

Short Answer

Expert verified
The short answers for each of the problems are as follows: a) The volume of the gas is approximately \(25.0\ L\). b) The absolute temperature of the gas is approximately \(1770.64\ K\). c) The pressure of the gas is approximately \(2.474 \times 10^6\ Pa\). d) The quantity of gas is approximately \(5.477\ mol\).

Step by step solution

01

(Problem a: Calculate volume)

First, we need to convert the temperature from Celsius to Kelvin. The formula to convert Celsius to Kelvin is as follows: \[T(K) =T(°C) + 273.15\] Given: - \(n = 1.50\) mol - \(P = 126.7\) kPa - \(T = -6^{\circ} C\) Now, we'll convert temperature and pressure to Kelvin and Pascal, respectively: \[T = -6 + 273.15 = 267.15 K\] \[P = 126.7 \times 10^3 Pa\] Next, we can solve for volume using the Ideal Gas Law formula: \[V = \frac{nRT}{P}\] \[V = \frac{(1.50\ mol)(8.314\ J/(mol·K))(267.15\ K)}{126.7 \times 10^3\ Pa}\] \[V \approx 0.0250\ m^3\] Now, convert cubic meters to liters: \[ V = 0.0250 \times 10^3 L \] \[ V \approx 25.0\ L \]
02

(Problem b: Calculate absolute temperature)

Given: - \(n = 3.33 \times 10^{-3}\) mol - \(P = 99.99\) kPa - \(V = 478\) mL Convert volume to cubic meters and pressure to Pascals: \[V = 478 \times 10^{-6} m^3\] \[P = 99.99 \times 10^3 Pa\] Now, we can solve for absolute temperature using the Ideal Gas Law: \[T = \frac{PV}{nR}\] \[T = \frac{(99.99 \times 10^3\ Pa)(478 \times 10^{-6} m^3)}{(3.33 \times 10^{-3}\ mol)(8.314\ J/(mol·K))}\] \[T \approx 1770.64\ K\]
03

(Problem c: Calculate pressure)

Given: - \(n = 0.00245\) mol - \(V = 413\) mL - \(T = 138^{\circ} C\) Convert temperature to Kelvin and volume to cubic meters: \[T = 138 + 273.15 = 411.15 K\] \[V = 413 \times 10^{-6} m^3\] Now, we can solve for pressure using the Ideal Gas Law: \[P = \frac{nRT}{V}\] \[P = \frac{(0.00245\ mol)(8.314\ J/(mol·K))(411.15\ K)}{413 \times 10^{-6}\ m^3}\] \[P \approx 2.474 \times 10^6\ Pa\]
04

(Problem d: Calculate quantity of gas in moles)

Given: - \(V = 126.5\) L - \(T = 54^{\circ} C\) - \(P = 11.25\) kPa Convert temperature to Kelvin, volume to cubic meters, and pressure to Pascals: \[T = 54 + 273.15 = 327.15 K\] \[P = 11.25 \times 10^3 Pa\] \[V = 126.5 \times 10^{-3} m^3\] Now, we can solve for the number of moles using the Ideal Gas Law: \[n = \frac{PV}{RT}\] \[n = \frac{(11.25 \times 10^3\ Pa)(126.5 \times 10^{-3}\ m^3)}{(8.314\ J/(mol·K))(327.15\ K)}\] \[n \approx 5.477\ mol\]

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Most popular questions from this chapter

Ammonia and hydrogen chloride react to form solid ammonium chloride: $$\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)$$ Two \(2.00-\mathrm{L}\) flasks at \(25^{\circ} \mathrm{C}\) are connected by a valve, as shown in the drawing. One flask contains \(5.00 \mathrm{~g}\) of \(\mathrm{NH}_{3}(g),\) and the other contains \(5.00 \mathrm{~g}\) of \(\mathrm{HCl}(g) .\) When the valve is opened, the gases react until one is completely consumed. (a) Which gas will remain in the system after the reaction is complete? (b) What will be the final pressure of the system after the reaction is complete? (Neglect the volume of the ammonium chloride formed.) (c) What mass of ammonium chloride will be formed?

A person weighing \(75 \mathrm{~kg}\) is standing on a threelegged stool. The stool momentarily tilts so that the entire weight is on one foot. If the contact area of each foot is \(5.0 \mathrm{~cm}^{2}\), calculate the pressure exerted on the underlying surface in (a) bars, \((\mathbf{b})\) atmospheres, and \((\mathbf{c})\) pounds per square inch.

A quantity of \(\mathrm{N}_{2}\) gas originally held at \(531.96 \mathrm{kPa}\) pressure in a 1.00 - \(\mathrm{L}\) container at \(26^{\circ} \mathrm{C}\) is transferred to a \(12.5-\mathrm{L}\) container at \(20^{\circ} \mathrm{C}\). A quantity of \(\mathrm{O}_{2}\) gas originally at \(531.96 \mathrm{kPa}\) and \(26^{\circ} \mathrm{C}\) in a \(5.00-\mathrm{L}\) container is transferred to this same container. What is the total pressure in the new container?

Perform the following conversions: (a) 0.912 atm to torr, (b) 0.685 bar to kilopascals, (c) \(655 \mathrm{~mm} \mathrm{Hg}\) to atmospheres, (d) \(1.323 \times 10^{5}\) Pa to atmospheres, (e) 2.50 atm to psi.

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