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 $\mathrm{L}$ ?

Short 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 $413.05Kor139.9°C$. (d) The pressure of the gas in the new container is approximately $52,127kPa$.

Step-by-step solution

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

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

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#

Pressure and temperature at STP

We know STP condition as: P = 1 atm and T = 273.15 K.

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#

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 $413.05Kor139.9°C$. #d) Pressure of the gas transferred to another container#

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

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 $52,127kPa$.

Key concepts

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.

Understanding these concepts allows for predicting and manipulating how gases will react under different conditions, which is essential in fields ranging from chemistry to engineering.

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:$$P(\text{kPa})=P(\text{atm})\times 101.325$$

Conversion between units is crucial when using the Ideal Gas Law, as it requires consistent units for accurate calculations. For example, in the problem, pressure is initially given in kPa but needs to be in atm (or vice versa) to properly apply the gas constant $R$. This uniformity in units ensures the precision and accuracy of the results.

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:$$n=\frac{extmass}{extmolarmass}$$
• To find mass from moles:$$\text{mass}=n\times \text{molar mass}$$

Calculating the molar mass aids in determining how much of a gas is present in a given volume under specified conditions. Coupling molar mass with the Ideal Gas Law, we can find either the mass or the number of moles of gas and further solve for other properties like volume and pressure.

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:$$V=\frac{nRT}{P}$$

Here, $n$ is the number of moles, $R$ is the appropriate gas constant, and $T$ and $P$ are set to their STP values (273.15 K and 1 atm, respectively). This helps in understanding how gas volume changes under standard conditions, a crucial concept for both scientific calculations and practical applications.