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A flask contains \(201 \mathrm{~mL}\) of argon at \(21^{\circ} \mathrm{C}\) and \(738 \mathrm{mmHg}\). What is the volume of gas, corrected to STP?

Short Answer

Expert verified
The volume of gas at STP is approximately 182.3 mL.

Step by step solution

01

Understand STP Conditions

STP (Standard Temperature and Pressure) conditions are defined as a temperature of \(0^{\circ}\mathrm{C}\) (or \(273.15\, \mathrm{K}\)) and a pressure of \(1\, \mathrm{atm}\) (or \(760\, \mathrm{mmHg}\)). We need to convert the given volume of gas to these standard conditions.
02

Use the Combined Gas Law

The combined gas law relates pressure, volume, and temperature of a gas: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\), where \(P\) is pressure, \(V\) is volume, \(T\) is temperature (in Kelvin).
03

Convert Temperatures to Kelvin

Convert the given temperature from Celsius to Kelvin using the formula: \(T(\mathrm{K}) = T(^{\circ}C) + 273.15\). So, \(21^{\circ}C = 294.15\, \mathrm{K}\). The STP temperature is already in Kelvin: \(273.15\, \mathrm{K}\).
04

Set Up the Equations

Given: \(P_1 = 738\, \mathrm{mmHg}\), \(V_1 = 201\, \mathrm{mL}\), \(T_1 = 294.15\, \mathrm{K}\). At STP: \(P_2 = 760\, \mathrm{mmHg}\) and \(T_2 = 273.15\, \mathrm{K}\). We solve for \(V_2\): \(\frac{738 \times 201}{294.15} = \frac{760 \times V_2}{273.15}\).
05

Solve for V2

Rearrange and solve the equation: \(V_2 = \frac{738 \times 201 \times 273.15}{294.15 \times 760}\). Calculate \(V_2\) to find the volume of the gas at STP conditions.
06

Calculate the Final Volume

Perform the calculations: \(V_2 \approx \frac{738 \times 201 \times 273.15}{294.15 \times 760} \approx 182.3\). So, the corrected volume at STP is approximately \(182.3\, \mathrm{mL}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Temperature and Pressure (STP)
Standard Temperature and Pressure, commonly abbreviated as STP, are standard conditions used to ensure consistency in scientific calculations involving gases. It is easier to compare data when collected under the same conditions. At STP, the temperature is defined as 0°C, which is equivalent to 273.15 K. The pressure is 1 atmosphere, which can be represented in different units. Commonly, pressure at STP is noted as 1 atm or 760 mmHg.

These standard conditions are essential when using gas laws, as real-life measurements often differ. Converting data into STP allows for a uniform reference point. This simplification is particularly useful for comparative studies and theoretical calculations in chemistry and physics. Understanding STP allows for accuracy and simplifies solutions.
Conversion of units in gas laws
Gas laws, such as the combined gas law, often involve units for pressure and volume that need conversion to ensure calculations are accurate. In the problem at hand, pressure is initially given in mmHg, but you might also encounter it in atm or torr, each requiring specific conversions. For example, 1 atm is equivalent to 760 mmHg or 760 torr.

Volume measurements, such as milliliters or liters, may also need conversion depending on the formula used. Generally, care should be taken to ensure that all units in the gas law equations are consistent before calculations are performed.

Accurate unit conversion is crucial, as it lays the foundation for precise scientific work. It prevents errors in calculations and helps maintain the integrity of the results. Always double-check your unit conversions when solving problems related to gas laws.
Temperature conversion from Celsius to Kelvin
Temperature is a critical component in gas laws and must always be expressed in Kelvin for calculations. The Kelvin scale is an absolute temperature scale where 0 K represents absolute zero, the theoretical point of no molecular motion. This makes Kelvin ideal for equations involving proportional relationships, such as in the combined gas law.

To convert Celsius to Kelvin, add 273.15 to the Celsius temperature. For instance, the exercise shows a conversion from 21°C to 294.15 K. This conversion ensures that the temperature is effectively used in relevant formulas and calculations involving gases.

Always remember, without proper conversion of temperature units to Kelvin, results may be inconsistent or incorrect. Mastering this simple conversion can facilitate smoother, more accurate calculations in chemistry and related disciplines.

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Most popular questions from this chapter

In the presence of a platinum catalyst, ammonia, \(\mathrm{NH}_{3}\), burns in oxygen, \(\mathrm{O}_{2}\), to give nitric oxide, NO, and water vapor. How many volumes of nitric oxide are obtained from one volume of ammonia, assuming each gas is at the -same temperature and pressure?

You vaporize a liquid substance at \(100^{\circ} \mathrm{C}\) and \(755 \mathrm{mmHg}\). The volume of \(0.548 \mathrm{~g}\) of vapor is \(237 \mathrm{~mL}\). What is the molecular weight of the substance?

Methanol, \(\mathrm{CH}_{3} \mathrm{OH},\) can be produced in industrial plants by reacting carbon dioxide with hydrogen in the presence of a catalyst. Water is the other product. How many volumes of hydrogen are required for each volume of carbon dioxide when each gas is at the same temperature and pressure?

Calculate the rms speed of \(\mathrm{Br}_{2}\) molecules at \(23^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm} .\) What is the rms speed of \(\mathrm{Br}_{2}\) at \(23^{\circ} \mathrm{C}\) and \(2.00 \mathrm{~atm} ?\)

A 1.000 -g sample of an unknown gas at \(0^{\circ} \mathrm{C}\) gives the following data: \(P(a t m) \quad V(L)\) 0.2500 $$ 3.1908 $$ $$ \begin{array}{ll} 0.5000 & 1.5928 \\\ 0.7500 & 1.0601 \end{array} $$ \(\begin{array}{ll}1.0000 & 0.7930\end{array}\) Use these data to calculate the value of the molar mass at each of the given pressures from the ideal gas law (we will call this the "apparent molar mass" at this pressure). Plot the apparent molar masses against pressure and extrapolate to find the molar mass at zero pressure. Because the ideal gas law is most accurate at low pressures, this extrapolation will give an accurate value for the molar mass. What is the accurate molar mass?

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