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Chapter 10: Problem 39

A scuba diver's tank contains \(2.50 \mathrm{~kg}\) of \(\mathrm{O}_{2}\) compressed into a volume of \(11.0 \mathrm{~L}\). (a) Calculate the gas pressure inside the tank at \(10^{\circ} \mathrm{C}\). (b) What volume would this oxygen occupy at \(25^{\circ} \mathrm{C}\) and \(101.33 \mathrm{kPa} ?\)

Short Answer

Expert verified
The gas pressure inside the tank is 166.5 kPa, and the oxygen would occupy a volume of 1.90 L at \(25^{\circ}C\) and \(101.33\:kPa\).

Step by step solution

01

(Step 1) Converting given values to appropriate units.

Before we use the ideal gas law, we must convert the given values into appropriate units. The mass of oxygen is given in kg, which should be converted to moles (n) using the molar mass of oxygen which is approximately 32 g/mol. The temperature should be converted to Kelvin (K) for the ideal gas law calculation. \(mass\_of\_O2 = 2.50\:kg\) \(volume\_of\_tank = 11.0\:L\) To convert the mass of oxygen to moles : \(moles = \frac{mass}{molar\_mass}\) where \(molar\_mass\_of\_O2 = 32\:g/mol\)
02

(Step 2) Calculate the number of moles of oxygen

First, we need to convert the mass of oxygen from kg to grams (1 kg = 1000 g). \(mass\_of\_O2 = 2.50\:kg * 1000\:g/kg = 2500\:g\) Next, we will calculate the number of moles of oxygen (n) using the molar mass of oxygen. \(moles = \frac{mass\_of\_O2}{molar\_mass\_of\_O2}\) \(moles = \frac{2500\:g}{32\:g/mol} = 78.125 \:mol\)
03

(Step 3) Convert temperature to Kelvin

To use the ideal gas law, we need to convert the temperature from Celsius to Kelvin. \(Tem_1 = 10^{\circ}C\) \(Tem_2 = 25^{\circ}C\) Therefore, \(Tem_1\_K = Tem_1 + 273.15 = 283.15\:K\) \(Tem_2\_K = Tem_2 + 273.15 = 298.15\:K\)
04

(Step 4) Calculate the gas pressure inside the tank

Now we can use the ideal gas law to calculate the pressure inside the tank. The ideal gas law is: \(PV = nRT\) Rearrange for pressure (P): \(P = \frac{nRT}{V}\) where P = pressure V = volume n = moles R = ideal gas constant (8.314 J/mol K) T = temperature in Kelvin Plug in the values for the tank: \(P = \frac{78.125\:mol * 8.314\:J/mol\:K * 283.15\:K}{11.0\:L}\) \(P = \frac{183153.2\:J/K}{11.0\:L}\) To convert J/mol to kPa, we have to divide by 100 (1 kPa = 100 J/mol): Pressure \(P = \frac{183153.2\:J/K}{11.0\:L * 100}\) \(P = 166.5\:kPa\) So, the gas pressure inside the tank is 166.5 kPa.
05

(Step 5) Calculate the new volume of oxygen at given temperature and pressure

We will again use the ideal gas law to find the new volume of oxygen at \(25^{\circ}C\) and \(101.33\:kPa\). First, set up this equation: \(P_1V_1/T_1 = P_2V_2/T_2\) Rearrange to find \(V_2\): \(V_2 = \frac{P_1V_1 T_2}{P_2T_1}\) Plug in the values: \(V_2 = \frac{166.5\:kPa * 11.0\:L * 298.15\:K}{101.33\:kPa * 283.15\:K}\) \(V_2 = \frac{54617.925\:L\:K}{28683.3\: kPa\:K}\) \(V_2 = 1.90\:L\) So, the oxygen would occupy a volume of 1.90 L at \(25^{\circ}C\) and \(101.33\:kPa\).

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Most popular questions from this chapter

(a) Amonton's law expresses the relationship between pressure and temperature. Use Charles's law and Boyle's law to derive the proportionality relationship between \(P\) and \(T .(\mathbf{b})\) If a car tire is filled to a pressure of \(220.6 \mathrm{kPa}\) measured at \(24^{\circ} \mathrm{C}\), what will be the tire pressure if the tires heat up to \(49^{\circ} \mathrm{C}\) during driving?

A glass vessel fitted with a stopcock valve has a mass of $337.428 \mathrm{~g}\( when evacuated. When filled with \)\mathrm{Ar}$, it has a mass of \(339.854 \mathrm{~g}\). When evacuated and refilled with a mixture of Ne and Ar, under the same conditions of temperature and pressure, it has a mass of \(339.076 \mathrm{~g} .\) What is the mole percent of Ne in the gas mixture?

(a) What are the mole fractions of each component in a mixture of $15.08 \mathrm{~g}\( of \)\mathrm{O}_{2}, 8.17 \mathrm{~g}\( of \)\mathrm{N}_{2},$ and \(2.64 \mathrm{~g}\) of \(\mathrm{H}_{2}\) (b) What is the partial pressure in atm of each component of this mixture if it is held in a \(15.50-\mathrm{L}\) vessel at \(15^{\circ} \mathrm{C} ?\)

(a) What conditions are represented by the abbreviation STP? (b) What is the molar volume of an ideal gas at STP? (c) Room temperature is often assumed to be \(25^{\circ} \mathrm{C}\). Calculate the molar volume of an ideal gas at \(25^{\circ} \mathrm{C}\) and \(101.3 \mathrm{kPa}\) pressure. \((\mathbf{d})\) If you measure pressure in bars instead of atmospheres, calculate the corresponding value of \(R\) in \(\mathrm{L}\) -bar \(/ \mathrm{mol}-\mathrm{K}\).

The atmospheric concentration of \(\mathrm{CO}_{2}\) gas is presently 407 ppm (parts per million, by volume; that is, \(407 \mathrm{~L}\) of every $10^{6} \mathrm{~L}\( of the atmosphere are \)\mathrm{CO}_{2}$ ). What is the mole fraction of \(\mathrm{CO}_{2}\) in the atmosphere?
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