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(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 1 atm pressure.

Short Answer

Expert verified
(a) STP stands for Standard Temperature and Pressure, which are 0°C (273.15 K) and 1 atm (101.325 kPa or 760 mmHg). (b) The molar volume of an ideal gas at STP is 22.41 L/mol. (c) The molar volume of an ideal gas at 25°C and 1 atm pressure is 24.47 L/mol.

Step by step solution

01

(a) Definition of STP

STP stands for Standard Temperature and Pressure. It is a reference condition used in chemistry, physics, and engineering to simplify calculations involving gases. The conditions represented by STP are: - Temperature: 0°C (273.15 K) - Pressure: 1 atm (101.325 kPa or 760 mmHg)
02

(b) Molar volume of an ideal gas at STP

The ideal gas law formula is given by: \(PV=nRT\), where P is pressure, V is volume, n is the amount of gas in moles, R is the ideal gas constant, and T is temperature. At STP, we want to find the molar volume (volume per mole) of an ideal gas. We will use Avogadro's law, which states that equal volumes of gases at the same temperature and pressure contain an equal number of particles (moles). One mole of an ideal gas at STP occupies a volume called the molar volume. For an ideal gas: Molar volume (V) = \( \frac{RT}{P} \) For STP conditions, temperature T = 273.15 K and pressure P = 1 atm. The ideal gas constant R = 0.0821 L atm / K mol Now, we can calculate the molar volume: Molar volume (V) = \( \frac{(0.0821 \, L \, atm / K \,mol) (273.15 \, K)}{1 \, atm} \) Molar volume (V) = 22.41 L/mol So, the molar volume of an ideal gas at STP is 22.41 L/mol.
03

(c) Molar volume of an ideal gas at 25°C and 1 atm pressure

Now, we need to calculate the molar volume of an ideal gas at room temperature, which is given as 25°C (or 298.15 K), and at 1 atm pressure. We will use the same formula as in part (b), but this time with the temperature T = 298.15 K and the same pressure P = 1 atm : Molar volume (V) = \( \frac{RT}{P} \) Molar volume (V) = \( \frac{(0.0821\, L\, atm / K\, mol)(298.15\, K)}{1\, atm} \) Molar volume (V) = 24.47 L/mol So, the molar volume of an ideal gas at 25°C (room temperature) and 1 atm pressure is 24.47 L/mol.

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Key Concepts

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

Ideal Gas Law
The ideal gas law is an essential equation in chemistry that helps us understand the behavior of gases. It is expressed as \(PV = nRT\), where:
  • \(P\) stands for pressure measured in atmospheres (atm).
  • \(V\) is the volume of the gas in liters (L).
  • \(n\) represents the number of moles of gas.
  • \(R\) is the ideal gas constant, which typically has a value of 0.0821 L atm / K mol.
  • \(T\) is the temperature in Kelvin (K).

This equation helps illustrate the relationship between the properties of an ideal gas. If you know any three of the variables, you can calculate the fourth. By applying this formula, we can determine the molar volume of gases under different conditions including STP and room temperature.
STP Conditions
STP, or Standard Temperature and Pressure, is a baseline set of conditions for comparing gas reactions across various scientific fields. These standardized conditions are crucial for making calculations simpler and ensuring consistency in experiments.
At STP, the temperature is precisely 0°C (which also translates to 273.15 K), and the pressure is 1 atm (or equivalently 101.325 kPa).
Under these conditions, one mole of an ideal gas occupies a volume known as the molar volume, which we can calculate using the ideal gas law formula. The formula for molar volume at STP is:\[V = \frac{RT}{P}\]Substituting in the values specific to STP conditions, with \(R = 0.0821 \text{ L atm / K mol}\), \(T = 273.15 \text{ K}\), and \(P = 1 \text{ atm}\), results in a molar volume of 22.41 L/mol. This means every mole of an ideal gas at STP takes up 22.41 liters of space.
Room Temperature Calculations
Room temperature is commonly accepted as approximately 25°C, which equals 298.15 K. Chemistry problems often involve calculating gas properties at this temperature because it reflects many real-world laboratory and environmental conditions.

To calculate the molar volume of an ideal gas at room temperature (25°C) and 1 atm pressure, we apply the ideal gas law equation as follows:\[Molar \ Volume \, (V) = \frac{RT}{P} \]Here, \(T\) is updated to 298.15 K, while \(P\) and \(R\) remain 1 atm and 0.0821 L atm / K mol, respectively. Plugging these values into the formula gives:\[Molar \ Volume \, (V) = \frac{(0.0821 \times 298.15)}{1} \approx 24.47 \text{ L/mol}\]As a result, at room temperature and 1 atm pressure, one mole of an ideal gas occupies about 24.47 liters. This slight increase from the STP molar volume reflects the impact of higher temperatures on gas expansion.

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Most popular questions from this chapter

(a) Place the following gases in order of increasing average molecular speed at \(25^{\circ} \mathrm{C}: \mathrm{Ne}, \mathrm{HBr}, \mathrm{SO}_{2}, \mathrm{NF}_{3}, \mathrm{CO}\) (b) Calculate the rms speed of \(\mathrm{NF}_{3}\) molecules at \(25^{\circ} \mathrm{C} .\) (c) Calculate the most probable speed of an ozone molecule in the stratosphere, where the temperature is \(270 \mathrm{~K}\).

(a) If the pressure exerted by ozone, \(\mathrm{O}_{3}\), in the stratosphere is \(3.0 \times 10^{-3}\) atm and the temperature is \(250 \mathrm{~K}\), how many ozone molecules are in a liter? (b) Carbon dioxide makes up approximately \(0.04 \%\) of Earth's atmosphere. If you collect a 2.0 - \(\mathrm{L}\) sample from the atmosphere at sea level \((1.00\) atm \()\) on a warm day \(\left(27^{\circ} \mathrm{C}\right),\) how many \(\mathrm{CO}_{2}\) molecules are in your sample?

In Sample Exercise 10.16 , we found that one mole of \(\mathrm{Cl}_{2}\) confined to \(22.41 \mathrm{~L}\) at \(0{ }^{\circ} \mathrm{C}\) deviated slightly from ideal behavior. Calculate the pressure exerted by \(1.00 \mathrm{~mol} \mathrm{Cl}_{2}\) confined to a smaller volume, \(5.00 \mathrm{~L}\), at \(25^{\circ} \mathrm{C} .\) (a) First use the ideal-gas equation and (b) then use the van der Waals equation for your calculation. (Values for the van der Waals constants are given in Table \(10.3 .)\) (c) Why is the difference between the result for an ideal gas and that calculated using the van der Waals equation greater when the gas is confined to \(5.00 \mathrm{~L}\) compared to \(22.4 \mathrm{~L} ?\)

How does a gas compare with a liquid for each of the following properties: (a) density, (b) compressibility, (c) ability to mix with other substances of the same phase to form homogeneous mixtures, \((\mathrm{d})\) ability to conform to the shape of its container?

Chlorine dioxide gas \(\left(\mathrm{ClO}_{2}\right)\) is used as a commercial bleaching agent. It bleaches materials by oxidizing them. In the course of these reactions, the \(\mathrm{ClO}_{2}\) is itself reduced. (a) What is the Lewis structure for \(\mathrm{ClO}_{2} ?\) (b) Why do you think that \(\mathrm{ClO}_{2}\) is reduced so readily? (c) When a \(\mathrm{ClO}_{2}\) molecule gains an electron, the chlorite ion, \(\mathrm{ClO}_{2}^{-}\), forms. Draw the Lewis structure for \(\mathrm{ClO}_{2}^{-}\). (d) Predict the \(\mathrm{O}-\mathrm{Cl}-\mathrm{O}\) bond angle in the \(\mathrm{ClO}_{2}^{-}\) ion. (e) One method of preparing \(\mathrm{ClO}_{2}\) is by the reaction of chlorine and sodium chlorite: $$ \mathrm{Cl}_{2}(g)+2 \mathrm{NaClO}_{2}(s) \longrightarrow 2 \mathrm{ClO}_{2}(g)+2 \mathrm{NaCl}(s) $$ If you allow \(15.0 \mathrm{~g}\) of \(\mathrm{NaClO}_{2}\) to react with \(2.00 \mathrm{~L}\) of chlorine gas at a pressure of 1.50 atm at \(21^{\circ} \mathrm{C}\), how many grams of \(\mathrm{ClO}_{2}\) can be prepared?

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