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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: \[T(K) =T(掳C) + 273.15\] 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: \[V = \frac{nRT}{P}\] \[V = \frac{(1.50\ mol)(8.314\ J/(mol路K))(267.15\ K)}{126.7 \times 10^3\ Pa}\] \[V \approx 0.0250\ m^3\] 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: \[T = \frac{PV}{nR}\] \[T = \frac{(99.99 \times 10^3\ Pa)(478 \times 10^{-6} m^3)}{(3.33 \times 10^{-3}\ mol)(8.314\ J/(mol路K))}\] \[T \approx 1770.64\ K\]
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: \[P = \frac{nRT}{V}\] \[P = \frac{(0.00245\ mol)(8.314\ J/(mol路K))(411.15\ K)}{413 \times 10^{-6}\ m^3}\] \[P \approx 2.474 \times 10^6\ Pa\]
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: \[n = \frac{PV}{RT}\] \[n = \frac{(11.25 \times 10^3\ Pa)(126.5 \times 10^{-3}\ m^3)}{(8.314\ J/(mol路K))(327.15\ K)}\] \[n \approx 5.477\ mol\]

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Key Concepts

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

Gas Volume Calculation
The Ideal Gas Law is a fundamental equation that helps us to calculate the volume of a gas under different conditions. When using this law, we follow the formula: \[ V = \frac{nRT}{P} \]Here:
  • \(V\) represents the volume in liters.
  • \(n\) is the amount of the gas in moles.
  • \(R\) is the ideal gas constant, which is \(8.314 \ \text{J/(mol}\cdot\text{K)}\).
  • \(T\) stands for temperature in Kelvin.
  • \(P\) is the pressure in Pascals.
To work with this equation, it is important to convert all units to their SI counterparts. This means:
  • Convert pressure from kPa to Pa by multiplying by 1000.
  • Convert the temperature from 掳C to Kelvin by adding 273.15.
  • Ensure volume is converted to cubic meters when necessary before calculating, and finally convert back to liters if needed (1000 L = 1 \(m^3\)).
Absolute Temperature Calculation
Calculating the absolute temperature of a gas involves using the Ideal Gas Law, but this time we solve the equation for temperature \(T\):\[ T = \frac{PV}{nR} \]Here are the steps to find the absolute temperature:
  • Ensure the pressure \(P\) is in Pascals and the volume \(V\) is in cubic meters.
  • The number of moles \(n\) and the ideal gas constant \(R\) remain the same as in other calculations.
A key factor in these calculations is ensuring all units are accurately converted before plugging values into the equation. After calculating, the result for \(T\) will be in Kelvin, the absolute temperature scale. This is important because Kelvin is an absolute measure where zero represents the absence of thermal energy.
Pressure Calculation
Pressure calculation with the Ideal Gas Law involves rearranging the formula to solve for pressure \(P\):\[ P = \frac{nRT}{V} \]When calculating:
  • Ensure the number of moles \(n\), the ideal gas constant \(R\), and the temperature \(T\) in Kelvin are well defined and known.
  • Convert the volume \(V\) to cubic meters for the calculation.
The results will give you pressure in Pascals (Pa). The importance of getting the conversion right is crucial, as pressure is sensitive to the changes in temperature, volume, and moles of the gas. This calculation helps in understanding the conditions under which gases operate and allows for predicting the behavior of gases under different applied conditions.
Mole Calculation
Calculating the number of moles \(n\) of a gas using the Ideal Gas Law is another essential application of this formula. Use:\[ n = \frac{PV}{RT} \]In this instance, to find \(n\), ensure:
  • The pressure \(P\) is in Pascals.
  • The volume \(V\) is in cubic meters.
  • The temperature \(T\) is in Kelvin.
  • The gas constant \(R\) remains at \(8.314 \ \text{J/(mol}\cdot\text{K)}\).
The result will tell you how many moles of the gas are present under the given conditions. This is critical in chemical reactions where amounts of reactants and products are often expressed in moles. Understanding this concept also aids in comprehending how gases relate to other states of matter in different physical and chemical processes.

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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?

Which of the following statements is false? (a) Gases are far less dense than liquids. (b) Gases are far more compressible than liquids. (c) Because liquid water and liquid carbon tetrachloride do not mix, neither do their vapors. (d) The volume occupied by a gas is determined by the volume of its container.

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required \(105 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of the gas to effuse. Under identical experimental conditions it required \(31 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of \(\mathrm{O}_{2}\) gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of \(1.0 \mathrm{~L} ;\) in other words, rate is the amount that diffuses over the time it takes to diffuse.)

Determine whether each of the following changes will increase, decrease, or not affect the rate with which gas molecules collide with the walls of their container: (a) increasing the volume of the container, \((\mathbf{b})\) increasing the temperature, (c) increasing the molar mass of the gas.

Table 10.3 shows that the van der Waals \(b\) parameter has units of \(\mathrm{L} / \mathrm{mol}\). This implies that we can calculate the size of atoms or molecules from \(b\). Using the value of \(b\) for \(\mathrm{Xe},\) calculate the radius of a Xe atom and compare it to the value found in Figure \(7.7,\) that is, \(140 \mathrm{pm}\). Recall that the volume of a sphere is \((4 / 3) \pi r^{3}\).
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