1. The pressure of a gas in a cylinder when heated to a temperature of 250 K is 1.5 atm. What is the initial temperature of the gas if its initial pressure was 1.00 atm? 2. List so... 1. The pressure of a gas in a cylinder when heated to a temperature of 250 K 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 cm^3 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, and what is the density of the gas at STP? Read more

Steps to Solve

Ex 3.6

1. Identify the Gas Law and Write the Formula

Since volume isn't mentioned and the problem involves pressure and temperature, we can apply Gay-Lussac's Law:

$$ \frac{P_1}{T_1} = \frac{P_2}{T_2} $$

2. List Known Variables

$P_1 = 1.00 , \text{atm}$

$T_2 = 250 , \text{K}$

$P_2 = 1.5 , \text{atm}$

3. Solve for $T_1$

Rearrange the formula to solve for $T_1$:

$T_1 = \frac{P_1 \cdot T_2}{P_2}$

4. Plug in values and solve

$T_1 = \frac{1.00 , \text{atm} \cdot 250 , \text{K}}{1.5 , \text{atm}}$

$T_1 \approx 166.67 , \text{K}$

Ex 3.6

2. Examples of Gay-Lussac's Law in everyday life

Examples include: * Pressure cooker: As temperature increases, pressure increases. * Aerosol cans: Heating an aerosol can may cause it to explode due to increased internal pressure. * Automobile tires: Tire pressure increases after driving due to the heat generated.

Ex 3.7

1. Identify the Gas Law and Write the Formula

Since we are given pressure, volume, and temperature changes, we should use the combined gas law:

$$ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} $$

2. List Known Variables

$P_1 = 60.0 , \text{kPa}$

$V_1 = 50 , \text{cm}^3$

$T_1 = 35^\circ\text{C} = 35 + 273.15 = 308.15 , \text{K}$

$P_2 = 1 , \text{atm} = 101.325 , \text{kPa}$

$T_2 = 0^\circ\text{C} = 273.15 , \text{K}$

We need to find $V_2$.

3. Solve for $V_2$

$V_2 = \frac{P_1V_1T_2}{P_2T_1}$

4. Plug in values and solve

$V_2 = \frac{60.0 , \text{kPa} \cdot 50 , \text{cm}^3 \cdot 273.15 , \text{K}}{101.325 , \text{kPa} \cdot 308.15 , \text{K}}$

$V_2 \approx 26.23 , \text{cm}^3 $

Ex 3.7

2. Identify variables and Gas Law We are given initial and final pressures, volumes and need to find the final Temperature. So, we are going to be using the combined gas law

   $$ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} $$

3. List Known Varibles

   $P_1 = 660 , \text{Torr}$

   $V_1 = 280 , \text{mL}$

   $T_1 = 26^\circ\text{C} = 26 + 273.15 = 299.15 , \text{K}$

   $P_2 = 940 , \text{Torr}$

   $V_2 = 440 , \text{mL}$

   We need to find $T_2$.

4. Solve for $T_2$

   $T_2 = \frac{P_2V_2T_1}{P_1V_1}$

5. Plug in values and solve for $T_2$

   $T_2 = \frac{940 , \text{Torr} \cdot 440 , \text{mL} \cdot 299.15 , \text{K}}{660 , \text{Torr} \cdot 280 , \text{mL}}$

   $T_2 \approx 672.47 , \text{K}$

6. Convert Kelvin to Celsius

   $T_2(^\circ\text{C}) = 672.47 - 273.15$

   $T_2 \approx 399.32^\circ\text{C}$

Ex 3.8

1. Find the Molecular Weight

Use the Ideal Gas Law to find the molar mass (molecular weight). First, write the Ideal Gas Law:

$PV = nRT$

Where:

$P$ = Pressure $V$ = Volume $n$ = Number of moles $R$ = Ideal gas constant $T$ = Temperature

We also know that density $\rho = \frac{m}{V}$, where $m$ is mass and $V$ is volume. We also know that $n = \frac{m}{M}$, where $M$ is the molar mass.

We can rearrange the Ideal Gas Law to solve for density and molar mass:

$PM = \rho R T$

$M = \frac{\rho R T}{P}$

However, we are not given the pressure or temperature. Instead, let's use the information given (1 mole of gas occupies 27.0 liters and its density is 1.41 g/L) directly.

Since we have 1 mole of gas and its volume and density, we can find the mass:

$m = \rho \cdot V = 1.41 , \text{g/L} \cdot 27.0 , \text{L} = 38.07 , \text{g}$

Since we have 1 mole, the molar mass is simply the mass we calculated:

$M = 38.07 , \text{g/mol}$

2. Calculate density at STP

STP (Standard Temperature and Pressure) is defined as $0^\circ \text{C}$ (273.15 K) and 1 atm.

Using the ideal gas law and molar mass we found earlier:

$\rho = \frac{PM}{RT}$

$\rho = \frac{1 , \text{atm} \cdot 38.07 , \text{g/mol}}{0.0821 , \text{L atm / (mol K)} \cdot 273.15 , \text{K}}$

$\rho \approx 1.697 , \text{g/L}$