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Chapter 10: Problem 41

A scuba diver's tank contains \(0.29 \mathrm{~kg}\) of \(\mathrm{O}_{2}\) compressed into a volume of 2.3 L. (a) Calculate the gas pressure inside the tank at \(9^{\circ} \mathrm{C} .\) (b) What volume would this oxygen occupy at \(26^{\circ} \mathrm{C}\) and 0.95 atm?

Short Answer

Expert verified
(a) The gas pressure inside the scuba diver's tank at \(9^\circ C\) is 10.15 atm. (b) At \(26^\circ C\) and 0.95 atm, the oxygen would occupy a volume of 29.4 L.

Step by step solution

01

Calculate the number of moles of Oxygen

First, we need to find the number of moles of Oxygen in the tank. To do this, we'll use the mass of Oxygen and its molar mass. The molar mass of O2 is 32 g/mol. Number of moles (n) = \(\frac{Mass}{Molar\ mass}\) n = \(\frac{0.29 kg}{32 g/mol}\) (1 kg = 1000 g) n = \(\frac{290 g}{32 g/mol}\) = \(9.06 mol\)
02

Convert temperature to Kelvin

We have to work with temperature in Kelvin while dealing with the gas law formula. To convert a Celsius temperature to Kelvin, we add 273.15 to the Celsius temperature. T1 (initial temperature) = \(9^\circ C + 273.15\) = \(282.15 K\) T2 (final temperature) = \(26^\circ C + 273.15\) = \(299.15 K\)
03

Calculate pressure inside the tank

Now, we'll use the ideal gas law formula to find the pressure inside the tank: \(PV = nRT\) We know the number of moles (n) and the initial temperature (T1). We also know the volume is 2.3 L, but we need to convert it to cubic meters (SI unit): V = 2.3 L × \(\frac{1 m^3}{1000 L}\) = \(2.3 × 10^{-3} m^3\) Now we can find the pressure: P = \(\frac{nRT}{V}\) Using R = \(8.314 J/(mol·K)\), we get: P = \(\frac{9.06 mol × 8.314 J/(mol·K) × 282.15 K}{2.3 × 10^{-3} m^3}\) = \(1028651.87 Pa\) We can convert the pressure to atm by using the conversion factor \(1 atm = 101325 Pa\): P = \(\frac{1028651.87 Pa}{101325 Pa/atm}\) = 10.15 atm
04

Calculate the final volume

We need to find the final volume of the oxygen gas at T2 and 0.95 atm. For this, we can use the combined gas law formula, which is derived from the ideal gas law and expresses the relationship between the initial and final states of a gas: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) We know \(P_1\), \(V_1\), \(T_1\), \(P_2\), and \(T_2\), so we can solve for \(V_2\): \(V_2 = \frac{P_1V_1T_2}{P_2T_1}\) \(V_2 = \frac{10.15 atm × 2.3 × 10^{-3} m^3 × 299.15 K}{0.95 atm × 282.15 K}\) \(V_2 = 0.0294 m^3\) Now, convert the volume back to liters: \(V_2 = 0.0294 m^3 × \frac{1000 L}{1 m^3}\) = 29.4 L The final volume of the oxygen at \(26^\circ C\) and 0.95 atm is 29.4 L.

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Most popular questions from this chapter

Calcium hydride, \(\mathrm{CaH}_{2}\), reacts with water to form hydrogen gas: $$ \mathrm{CaH}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+2 \mathrm{H}_{2}(g) $$ This reaction is sometimes used to inflate life rafts, weather balloons, and the like, when a simple, compact means of generating \(\mathrm{H}_{2}\) is desired. How many grams of \(\mathrm{CaH}_{2}\) are needed to generate \(145 \mathrm{~L}\) of \(\mathrm{H}_{2}\) gas if the pressure of \(\mathrm{H}_{2}\) is 825 torr at \(21{ }^{\circ} \mathrm{C}\) ?

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