1. The pressure of a gas at 250K is 1.5 atm. What is the initial temperature of the gas if its initial pressure was 1.00 atm? 2. List some examples of items that we use in our ever... 1. The pressure of a gas at 250K is 1.5 atm. What is the initial temperature of the gas if its initial pressure was 1.00 atm? 2. List some examples of items that we use in our everyday life that obey Gay-Lussac's law. 3. If a 50 cm³ sample of gas exerts a pressure of 60.0 kPa at 35°C, what volume will it occupy at STP (0°C and 1 atm)? 4. A 280 mL sample of neon exerts a pressure of 660 Torr at 26°C. At what temperature in °C would it exert a pressure of 940 Torr in a volume of 440 mL? 5. One mole of a gas occupies 27.0 liters. If its density is 1.41 g/L at a particular temperature and pressure, what is its molecular weight? What is the density of the gas at STP? Read more

Steps to Solve

1. Rephrasing Ex3.6, Question 1

A gas in a cylinder has a pressure of 1.5 atm at 250 K. If the initial pressure was 1.00 atm, what was the initial temperature? This problem can be sovled using Gay Lussac's Law.

2. Applying Gay-Lussac's Law

Gay-Lussac's Law states that $P_1/T_1 = P_2/T_2$, where $P_1$ and $T_1$ are the initial pressure and temperature, and $P_2$ and $T_2$ are the final pressure and temperature. We are given $P_1 = 1.00 \text{ atm}$, $P_2 = 1.5 \text{ atm}$, and $T_2 = 250 \text{ K}$. We want to find $T_1$.

3. Solving for $T_1$

Rearrange the formula to solve for $T_1$: $T_1 = \frac{P_1 \cdot T_2}{P_2}$. Now, plug in the given values: $T_1 = \frac{1.00 \text{ atm} \cdot 250 \text{ K}}{1.5 \text{ atm}} = \frac{250}{1.5} \text{ K} \approx 166.67 \text{ K}$

4. Rephrasing Ex3.6, Question 2

List some examples of items that use Gay-Lussac's law in everyday life.

5. Examples of items that obey Gay-Lussac's law

Examples include aerosol cans and pressure cookers.

6. Rephrasing Ex3.7, Question 1

If a 50 cm$^3$ sample of gas exerts a pressure of 60.0 kPa at 35$^\circ$C, what volume will it occupy at STP (0$^\circ$C & 1 atm)?

7. Converting Units and Applying the Combined Gas Law

First, convert the temperature to Kelvin: $T_1 = 35^\circ\text{C} + 273.15 = 308.15 \text{ K}$ and $T_2 = 0^\circ\text{C} + 273.15 = 273.15 \text{ K}$. Also, $P_2 = 1 \text{ atm} = 101.325 \text{ kPa}$.

The combined gas law is $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$.

8. Solving for $V_2$

We are given $V_1 = 50 \text{ cm}^3$, $P_1 = 60.0 \text{ kPa}$, $T_1 = 308.15 \text{ K}$, $P_2 = 101.325 \text{ kPa}$, and $T_2 = 273.15 \text{ K}$. We want to find $V_2$.

Rearrange the formula: $V_2 = \frac{P_1V_1T_2}{P_2T_1}$. Plug in the values: $V_2 = \frac{60.0 \text{ kPa} \cdot 50 \text{ cm}^3 \cdot 273.15 \text{ K}}{101.325 \text{ kPa} \cdot 308.15 \text{ K}} \approx 26.12 \text{ cm}^3$.

9. Rephrasing Ex3.7, Question 2

A 280 mL sample of neon exerts a pressure of 660 Torr at 26$^\circ$C. At what temperature in $^\circ$C would it exert a pressure of 940 Torr in a volume of 440 mL?

10. Applying the Combined Gas Law

We have $V_1 = 280 \text{ mL}$, $P_1 = 660 \text{ Torr}$, $T_1 = 26^\circ\text{C} + 273.15 = 299.15 \text{ K}$, $V_2 = 440 \text{ mL}$, and $P_2 = 940 \text{ Torr}$. We want to find $T_2$.

Using the combined gas law $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$.

11. Solving for $T_2$

Rearrange the formula: $T_2 = \frac{P_2V_2T_1}{P_1V_1}$. Plug in the values: $T_2 = \frac{940 \text{ Torr} \cdot 440 \text{ mL} \cdot 299.15 \text{ K}}{660 \text{ Torr} \cdot 280 \text{ mL}} \approx 672.78 \text{ K}$.

Convert back to Celsius: $T_2 = 672.78 \text{ K} - 273.15 = 399.63 ^\circ\text{C}$. Approximately $399.6^\circ\text{C}$.

12. Rephrasing Ex3.8, Question 1

One mole of a gas occupies 27.0 liters. If its density is 1.41 g/L at a particular temperature & pressure, what is its molecular weight?

13. Calculating Molecular Weight

Molecular weight (MW) can be calculated using the formula: $MW = \frac{\text{density} \cdot \text{volume}}{\text{moles}}$ We have density = 1.41 g/L, volume = 27.0 L, and moles = 1. $MW = \frac{1.41 \frac{\text{g}}{\text{L}} \cdot 27.0 \text{ L}}{1 \text{ mole}} = 38.07 \frac{\text{g}}{\text{mole}}$

14. Rephrasing Ex3.8, Question 2

What is the density of the gas at STP?

15. Calculating Density at STP

At STP, 1 mole of any gas occupies 22.4 L. We know the molecular weight is 38.07 g/mol

Density = $\frac{\text{mass}}{\text{volume}} = \frac{38.07 \text{ g}}{22.4 \text{ L}} = 1.70 \frac{\text{g}}{\text{L}}$.