Problem 44
Question
Many gases are shipped in high-pressure containers. Consider a steel tank whose volume is 55.0 gallons that contains \(\mathrm{O}_{2}\) gas at a pressure of \(16,500 \mathrm{kPa}\) at \(23^{\circ} \mathrm{C}\) . (a) What mass of \(\mathrm{O}_{2}\) does the tank contain? (b) What volume would the gas occupy at STP? (c) At what temperature would the pressure in the tank equal 150.0 atm? (d) What would be the pressure of the gas, in kPa, if it were transferred to a container at \(24^{\circ} \mathrm{C}\) whose volume is 55.0 \(\mathrm{L}\) ?
Step-by-Step Solution
Verified Answer
(a) The mass of O2 in the tank is approximately 30,751 g. (b) The volume of the gas at STP is approximately 21,582 L. (c) The temperature needed for the pressure to be 150 atm is approximately \(413.05 K \ or \ 139.9 °C \). (d) The pressure of the gas in the new container is approximately \(52,127 kPa\).
1Step 1: Convert pressure and temperature to appropriate units
The given pressure is in kPa, so we need to convert it to atm, and the temperature is in Celsius, convert it to Kelvin.
1 atm = 101.325 kPa, so: (16,500 kPa) * (1 atm / 101.325 kPa) = 162.93 atm
Temperature in Kelvin: 23 °C + 273.15 = 296.15 K
2Step 2: Use the Ideal Gas Law formula to calculate the moles of O2
PV = nRT => n = PV / (RT)
n = (162.93 atm * 55.0 gallons) / (0.0821 L*atm /mol*K * 296.15 K)
As 1 gallon = 3.78541 L,
n = (162.93 atm * 55.0 * 3.78541 L) / (0.0821 L*atm /mol*K * 296.15 K)
n = 960.96 mol of O2
3Step 3: Calculate the mass of O2
Since the molar mass of O2 is 32 g/mol, the mass of O2 in the tank is:
mass = n * molar mass
mass = 960.96 mol * 32 g/mol = 30750.72 g
So, the mass of O2 is approximately 30,751 g.
#b) Volume of the gas at STP#
4Step 1: Pressure and temperature at STP
We know STP condition as: P = 1 atm and T = 273.15 K.
5Step 2: Use the Ideal Gas Law formula to calculate the volume at STP
Let V' be the volume at STP, so:
PV = nRT => V' = (n R T) / P
V' = (960.96 mol * 0.0821 L*atm /mol*K * 273.15 K) / 1 atm
V' = 21,581.58 L
So, the volume of the gas at STP is approximately 21,582 L.
#c) Temperature for the pressure at 150 atm#
6Step 1: Use the Ideal Gas Law formula to calculate the temperature
Let T' be the temperature at which the pressure is 150 atm, so:
PV = nRT => T' = (PV) / (nR)
T' = (150 atm * 55.0 * 3.78541 L) / (960.96 mol * 0.0821 L*atm /mol*K)
T' = 413.05 K
So, the temperature needed for the pressure to be 150 atm is approximately \(413.05 K \ or \ 139.9 °C \).
#d) Pressure of the gas transferred to another container#
7Step 1: Convert the temperature to Kelvin
The gas is transferred to the new container at 24 °C, so:
T_new = 24 + 273.15 = 297.15 K
8Step 2: Use the Ideal Gas Law formula to calculate the new pressure
Let P_new be the pressure in kPa, so:
P_new = (n R T_new) / V_new
P_new = (960.96 mol * 8.314 kPa*L/mol*K * 297.15 K) / 55.0 L
P_new = 52,126.99 kPa
So, the pressure of the gas in the new container is approximately \(52,127 kPa\).
Key Concepts
Gas LawsPressure ConversionMolar Mass CalculationVolume at STP
Gas Laws
Gas laws are fundamental to understanding the behavior of gases under varying conditions. A primary equation used to study gases is the Ideal Gas Law: \[PV = nRT\]This equation relates pressure \(P\), volume \(V\), number of moles \(n\), the gas constant \(R\), and temperature \(T\) in Kelvin. The Ideal Gas Law is an impeccable tool for solving a variety of gas-related problems like those concerning gas pressure, volume, and temperature changes.
- Pressure (P): It's the force that the gas molecules exert on the walls of their container, commonly measured in atmospheres (atm) or kilopascals (kPa).
- Volume (V): This is the space that the gas occupies, often measured in liters (L).
- Moles (n): It signifies the quantity of gas molecules, where each mole equals Avogadro's number of molecules.
- Gas constant (R): A constant that makes the units of the equation consistent, usually 0.0821 L·atm/mol·K or 8.314 kPa·L/mol·K.
- Temperature (T): Typically measured in Kelvin, this reflects the kinetic energy of the particles.
Pressure Conversion
In many problems involving gas laws, pressure needs to be converted between different units, such as from kilopascals (kPa) to atmospheres (atm) or vice versa. One atmosphere is equivalent to 101.325 kPa. To convert:
- From kPa to atm: \[P(\text{atm}) = \frac{P(\text{kPa})}{101.325}\]
- From atm to kPa:\[P(\text{kPa}) = P(\text{atm}) \times 101.325\]
Molar Mass Calculation
The molar mass calculation is significant when dealing with gases, as it links the number of moles to the mass of the gas. Molar mass is the mass of one mole of a substance and is usually expressed in grams per mole (g/mol). For oxygen (\(O_2\), which is a diatomic molecule), the molar mass is 32 g/mol.In the context of gas laws:
- To find moles from mass:\[n = \frac{ ext{mass}}{ ext{molar mass}}\]
- To find mass from moles:\[\text{mass} = n \times \text{molar mass}\]
Volume at STP
Standard Temperature and Pressure (STP) are conditions used as reference points in the study of gases. STP denotes a temperature of 273.15 K (0 °C) and a pressure of 1 atm. At these conditions, it is widely accepted that 1 mole of any ideal gas occupies approximately 22.4 liters.The volume of a gas at STP can be calculated using the Ideal Gas Law, where the standard condition values are adopted:
- Calculate the volume at STP using:\[V = \frac{nRT}{P}\]
Other exercises in this chapter
Problem 41
A 35.1 g sample of solid \(\mathrm{CO}_{2}(\) dry ice \()\) is added to a container at a temperature of 100 \(\mathrm{K}\) with a volume of 4.0 \(\mathrm{L} .\)
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