Ex 3.6 1. The pressure of a gas in a cylinder when heated to a temperature of 250K is 1.5 atm. What is the initial temperature of the gas if its initial pressure was 1.00 atm? 2. L... Ex 3.6 1. The pressure of a gas in a cylinder when heated to a temperature of 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. Ex 3.1 1. If a 50cm³ sample of gas exerts a pressure of 60.0 kPa at 35°C, what volume will it occupy at STP (0°C & 1 atm)? 2. A 280mL 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 440mL? Read more

Steps to Solve

1. Problem 1 from Ex 3.6: Identify the gas law and relevant values

This problem involves the relationship between pressure and temperature of a gas at constant volume, which is described by Gay-Lussac's Law. The formula for Gay-Lussac's Law is $P_1/T_1 = P_2/T_2$.

We are given: $P_2 = 1.5 \text{ atm}$ $T_2 = 250 \text{ K}$ $P_1 = 1.00 \text{ atm}$ We need to find $T_1$.

2. Apply Gay-Lussac's Law

Using the formula $P_1/T_1 = P_2/T_2$, we can solve for $T_1$: $T_1 = (P_1 \times T_2) / P_2$

3. Calculate $T_1$

$T_1 = (1.00 \text{ atm} \times 250 \text{ K}) / 1.5 \text{ atm} = 166.67 \text{ K}$

4. Problem 1 from Ex 3.1: Identify the appropriate gas law

This problem deals with the change in volume of a gas under different pressure and temperature conditions. We need to determine the volume at Standard Temperature and Pressure (STP). We can use the combined gas law:

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

We are given: $V_1 = 50 \text{ cm}^3$ $P_1 = 60.0 \text{ kPa}$ $T_1 = 35^\circ\text{C} = 35 + 273.15 = 308.15 \text{ K}$ STP conditions are: $P_2 = 1 \text{ atm} = 101.325 \text{ kPa}$ $T_2 = 0^\circ\text{C} = 273.15 \text{ K}$ We need to find $V_2$.

5. Rearrange and solve for $V_2$

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

6. Plug in the values and calculate $V_2$

$V_2 = \frac{(60.0 \text{ kPa}) \times (50 \text{ cm}^3) \times (273.15 \text{ K})}{(101.325 \text{ kPa}) \times (308.15 \text{ K})} = 26.1 \text{ cm}^3$

7. Problem 2 from Ex 3.1: Identify the gas law

This problem involves changes in pressure, volume, and temperature. Therefore, we use the combined gas law.

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

We are given: $V_1 = 280 \text{ mL}$ $P_1 = 660 \text{ Torr}$ $T_1 = 26^\circ\text{C} = 26 + 273.15 = 299.15 \text{ K}$ $P_2 = 940 \text{ Torr}$ $V_2 = 440 \text{ mL}$ We need to find $T_2$ in Celsius.

8. Solve for $T_2$

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

9. Plug in the values and calculate $T_2$ (in Kelvin)

$T_2 = \frac{(940 \text{ Torr}) \times (440 \text{ mL}) \times (299.15 \text{ K})}{(660 \text{ Torr}) \times (280 \text{ mL})} = 672.5 \text{ K}$

10. Convert $T_2$ from Kelvin to Celsius

$T_2 (\text{in }^\circ\text{C}) = T_2 (\text{in K}) - 273.15 = 672.5 - 273.15 = 399.35^\circ\text{C}$