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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 mol has a pressure of 1.25 atm 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 750 torr; (c) the pressure, in atmospheres, 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 \(\mathrm{L}\) at \(54^{\circ} \mathrm{C}\) has a pressure of 11.25 \(\mathrm{kPa}\) .

Short Answer

Expert verified
The short answer for each part is as follows: (a) The volume of the gas is \(16.54 L\). (b) The absolute temperature of the gas is \(170.90 K\). (c) The pressure of the gas is \(0.487 atm\). (d) The quantity of gas is \(5.92 mol\).

Step by step solution

01

Part (a): Calculate the volume of the gas

We are given the pressure \(P = 1.25 atm\), the number of moles \(n = 1.50 mol\), and the temperature \(T = -6^{\circ}C\). We have to find the volume of the gas (V). First, we need to convert the temperature into Kelvin (K). Temperature in Kelvin (K) = Temperature in Celsius (C) + 273.15 \(T = -6^{\circ}C + 273.15 = 267.15K\) Now we can use the ideal gas law equation \(PV = nRT\) to find the volume. Here, we will use the value of R in atm L/mol K which is 0.08206 \(atm L / (mol K)\). \(V = \frac{nRT}{P} = \frac{1.50 mol × 0.08206 \frac{atm L}{mol K} × 267.15K}{1.25 atm} = 16.54L\) The volume of the gas is 16.54 liters.
02

Part (b): Calculate the absolute temperature of the gas

We are given the number of moles \(n = 3.33 \times 10^{-3} mol\), the volume V = 478 mL, and the pressure \(P = 750 torr\). We need to find the temperature of the gas (T). First, we need to convert the pressure to atm and the volume to L: \(P = \frac{750 torr}{760 torr/atm} = 0.9868 atm\) \(V = \frac{478mL}{1000mL/L} = 0.478L\) Now we can use the ideal gas law equation \(PV = nRT\) to find the temperature. \(T = \frac{PV}{nR} = \frac{0.9868 atm × 0.478L}{3.33 \times 10^{-3} mol × 0.08206 \frac{atm L}{mol K}} = 170.90 K\) The absolute temperature of the gas is 170.90 K.
03

Part (c): Calculate the pressure of the gas

We are given the number of moles \(n = 0.00245 mol\), the volume \(V = 413 mL\), and the temperature \(T = 138^{\circ}C\). We have to find the pressure of the gas (P). First, convert the temperature to Kelvin and the volume to L: \(T = 138^{\circ}C + 273.15 = 411.15K\) \(V = \frac{413 mL}{1000 mL/L} = 0.413 L\) Now we can use the ideal gas law equation \(PV = nRT\) to find the pressure. \(P = \frac{nRT}{V} = \frac{0.00245 mol × 0.08206 \frac{atm L}{mol K} × 411.15K}{0.413 L} = 0.487 atm\) The pressure of the gas is 0.487 atmospheres.
04

Part (d): Calculate the quantity of gas in moles

We are given the volume \(V = 126.5L\), the temperature \(T = 54^{\circ}C\), and the pressure \(P = 11.25 kPa\). We have to find the number of moles (n) of the gas. First, we need to convert the temperature to Kelvin and the pressure to atm: \(T = 54^{\circ}C + 273.15 = 327.15 K\) \(P = \frac{11.25kPa}{101.325kPa/atm} = 0.1110 atm\) Now we can use the ideal gas law equation \(PV = nRT\) to find the number of moles. \(n = \frac{PV}{RT} = \frac{0.1110 atm × 126.5L}{0.08206 \frac{atm L}{mol K} × 327.15K} = 5.92 mol\) The quantity of gas is 5.92 moles.

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