Esercizi sulle Leggi dei Gas

Esercizi sulle Leggi dei Gas

Esercizi sulle Leggi dei Gas

Versione italiana

Esercizi sulle Leggi dei Gas

Introduzione

Le leggi dei gas descrivono il comportamento dei gas in relazione a pressione, volume e temperatura. Le leggi principali includono la legge di Boyle, la legge di Charles e la legge di Avogadro.

Esercizio 1: Legge di Boyle

Domanda: Un gas occupa un volume di 5 L a una pressione di 2 atm. Quale sarà il volume del gas se la pressione aumenta a 4 atm, mantenendo la temperatura costante?

Formula:

P_1 V_1 = P_2 V_2 

$$P_1 V_1 = P_2 V_2$$

Dati:

• P_1 = 2 \, \text{atm}$P_1 = 2 \, \text{atm}$
• V_1 = 5 \, \text{L}$V_1 = 5 \, \text{L}$
• P_2 = 4 \, \text{atm}$P_2 = 4 \, \text{atm}$

Risposta:

2 \, \text{atm} \times 5 \, \text{L} = 4 \, \text{atm} \times V_2

$$2 \, \text{atm} \times 5 \, \text{L} = 4 \, \text{atm} \times V_2$$

10 = 4 V_2 \implies V_2 = \frac{10}{4} = 2.5 \, \text{L}

$$10 = 4 V_2 \implies V_2 = \frac{10}{4} = 2.5 \, \text{L}$$

Esercizio 2: Legge di Charles

Domanda: Un gas occupa un volume di 10 L a 20 °C. Quale sarà il volume del gas se la temperatura aumenta a 60 °C, mantenendo la pressione costante?

Formula:

\frac{V_1}{T_1} = \frac{V_2}{T_2} 

$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$

Dati:

• V_1 = 10 \, \text{L}$V_1 = 10 \, \text{L}$
• T_1 = 20 \, \text{°C} = 293 \, \text{K}$T_1 = 20 \, \text{°C} = 293 \, \text{K}$ (convertito in Kelvin)
• T_2 = 60 \, \text{°C} = 333 \, \text{K}$T_2 = 60 \, \text{°C} = 333 \, \text{K}$

Risposta:

\frac{10 \, \text{L}}{293 \, \text{K}} = \frac{V_2}{333 \, \text{K}}

$$\frac{10 \, \text{L}}{293 \, \text{K}} = \frac{V_2}{333 \, \text{K}}$$

V_2 = \frac{10 \, \text{L} \times 333 \, \text{K}}{293 \, \text{K}} \approx 11.37 \, \text{L}

$$V_2 = \frac{10 \, \text{L} \times 333 \, \text{K}}{293 \, \text{K}} \approx 11.37 \, \text{L}$$

Esercizio 3: Legge di Avogadro

Domanda: Qual è il volume occupato da 2 mol di un gas ideale a 0 °C e 1 atm di pressione?

Formula:

V = n \times V_m 

$$V = n \times V_m$$

dove V_m$V_m$ è il volume molare di un gas ideale a condizioni standard (0 °C e 1 atm), che è circa 22.4 L/mol.

Dati:

• n = 2 \, \text{mol}$n = 2 \, \text{mol}$
• V_m = 22.4 \, \text{L/mol}$V_m = 22.4 \, \text{L/mol}$

Risposta:

V = 2 \, \text{mol} \times 22.4 \, \text{L/mol} = 44.8 \, \text{L}

$$V = 2 \, \text{mol} \times 22.4 \, \text{L/mol} = 44.8 \, \text{L}$$

Esercizio 4: Equazione dei gas ideali

Domanda: Un gas ideale ha una pressione di 3 atm, un volume di 10 L e una temperatura di 300 K. Quanti moli di gas ci sono?

Formula:

PV = nRT

$$PV = nRT$$

dove R = 0.0821 \, \text{L} \cdot \text{atm} / (\text{K} \cdot \text{mol})$R = 0.0821 \, \text{L} \cdot \text{atm} / (\text{K} \cdot \text{mol})$

Dati:

• P = 3 \, \text{atm}$P = 3 \, \text{atm}$
• V = 10 \, \text{L}$V = 10 \, \text{L}$
• T = 300 \, \text{K}$T = 300 \, \text{K}$

Risposta:

n = \frac{PV}{RT} = \frac{3 \, \text{atm} \times 10 \, \text{L}}{0.0821 \, \text{L} \cdot \text{atm} / (\text{K} \cdot \text{mol}) \times 300 \, \text{K}}

$$n = \frac{PV}{RT} = \frac{3 \, \text{atm} \times 10 \, \text{L}}{0.0821 \, \text{L} \cdot \text{atm} / (\text{K} \cdot \text{mol}) \times 300 \, \text{K}}$$

n \approx \frac{30}{24.63} \approx 1.22 \, \text{mol}

$$n \approx \frac{30}{24.63} \approx 1.22 \, \text{mol}$$

English version

Gas Laws Exercises

Introduction

Gas laws describe the behavior of gases with respect to pressure, volume, and temperature. The main laws include Boyle's Law, Charles's Law, and Avogadro's Law.

Exercise 1: Boyle's Law

Question: A gas occupies a volume of 5 L at a pressure of 2 atm. What will be the volume of the gas if the pressure is increased to 4 atm, keeping the temperature constant?

Formula:

P_1 V_1 = P_2 V_2 

$$P_1 V_1 = P_2 V_2$$

Data:

• P_1 = 2 \, \text{atm}$P_1 = 2 \, \text{atm}$
• V_1 = 5 \, \text{L}$V_1 = 5 \, \text{L}$
• P_2 = 4 \, \text{atm}$P_2 = 4 \, \text{atm}$

Answer:

2 \, \text{atm} \times 5 \, \text{L} = 4 \, \text{atm} \times V_2

$$2 \, \text{atm} \times 5 \, \text{L} = 4 \, \text{atm} \times V_2$$

10 = 4 V_2 \implies V_2 = \frac{10}{4} = 2.5 \, \text{L}

$$10 = 4 V_2 \implies V_2 = \frac{10}{4} = 2.5 \, \text{L}$$

Exercise 2: Charles's Law

Question: A gas occupies a volume of 10 L at 20 °C. What will be the volume of the gas if the temperature increases to 60 °C, keeping the pressure constant?

Formula:

\frac{V_1}{T_1} = \frac{V_2}{T_2} 

$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$

Data:

• V_1 = 10 \, \text{L}$V_1 = 10 \, \text{L}$ - T_1 = 20 \, \text{C°} = 293 \, \text{K}$T_1 = 20 \, \text{C°} = 293 \, \text{K}$ (converted to Kelvin)
• T_2 = 60 \, \text{°C} = 333 \,$T_2 = 60 \, \text{°C} = 333 \,$
  Answer:

\frac{10 \, \text{L}}{293 \, \text{K}} = \frac{V_2}{333 \, \text{K}}

$$\frac{10 \, \text{L}}{293 \, \text{K}} = \frac{V_2}{333 \, \text{K}}$$

V_2 = \frac{10 \, \text{L} \times 333 \, \text{K}}{293 \, \text{K}} \approx 11.37 \, \text{L}

$$V_2 = \frac{10 \, \text{L} \times 333 \, \text{K}}{293 \, \text{K}} \approx 11.37 \, \text{L}$$

Exercise 3: Avogadro's Law

Question: What is the volume occupied by 2 mol of an ideal gas at 0 °C and 1 atm of pressure?

Formula:

V = n \times V_m 

$$V = n \times V_m$$

where V_m$V_m$ is the molar volume of an ideal gas at standard conditions (0 °C and 1 atm), which is about 22.4 L/mol.

Data:

• n = 2 \, \text{mol}$n = 2 \, \text{mol}$
• V_m = 22.4 \, \text{L/mol}$V_m = 22.4 \, \text{L/mol}$

Answer:

V = 2 \, \text{mol} \times 22.4 \, \text{L/mol} = 44.8 \, \text{L}

$$V = 2 \, \text{mol} \times 22.4 \, \text{L/mol} = 44.8 \, \text{L}$$

Exercise 4: Ideal Gas Equation

Question: An ideal gas has a pressure of 3 atm, a volume of 10 L, and a temperature of 300 K. How many moles of gas are there?

Formula:

PV = nRT

$$PV = nRT$$

where
R = 0.0821 \, \text{L} \cdot \text{atm} / (\text{K} \cdot \text{mol})$R = 0.0821 \, \text{L} \cdot \text{atm} / (\text{K} \cdot \text{mol})$
Data:
P = 3 \, \text{atm}$P = 3 \, \text{atm}$ - V = 10 \, \text{L}$V = 10 \, \text{L}$ - T = 300 \, \text{K}$T = 300 \, \text{K}$ Answer:

n = \frac {PV}{RT} = \frac{3 \, \text{atm} \times 10 \, \text{L}}{0.0821 \, \text{L} \cdot \text{atm} / (\text{K} \cdot \text{mol}) \times 300 \, \text{K}}

$$n = \frac {PV}{RT} = \frac{3 \, \text{atm} \times 10 \, \text{L}}{0.0821 \, \text{L} \cdot \text{atm} / (\text{K} \cdot \text{mol}) \times 300 \, \text{K}}$$

n \approx \frac{30}{24.63} \approx 1.22 \, \text{mol}

$$n \approx \frac{30}{24.63} \approx 1.22 \, \text{mol}$$