Problem 28
Question
A fixed quantity of gas at \(21^{\circ} \mathrm{C}\) exhibits a pressure of 752 torr and occupies a volume of 5.12 L. (a) Calculate the volume the gas will occupy if the pressure is increased to 1.88 atm while the temperature is held constant. (b) Calculate the volume the gas will occupy if the temperature is increased to \(175^{\circ} \mathrm{C}\) while the pressure is held constant.
Step-by-Step Solution
Verified Answer
The gas will occupy approximately 2.69 L when the pressure is increased to 1.88 atm with constant temperature, and it will occupy approximately 7.84 L when the temperature is increased to 175°C with constant pressure.
1Step 1: (Part a: Using Boyle's Law to find the new volume)
To convert the initial pressure from torr to atm, use the conversion factor 1 atm = 760 torr:
P1 = (752 torr) * (1 atm / 760 torr) = 0.990 atm
Now let's apply Boyle's Law (P1V1 = P2V2):
\(0.990 \,\text{atm} \cdot 5.12 \,\text{L} = 1.88 \,\text{atm} \cdot V_2\)
Solving for \(V_2\), we get:
\(V_2 = \frac{0.990 \,\text{atm} \cdot 5.12 \,\text{L}}{1.88 \,\text{atm}} \approx 2.69 \,\text{L}\)
So, the volume of the gas will be approximately 2.69 L when the pressure is increased to 1.88 atm while the temperature is held constant.
2Step 2: (Part b: Using Charles' Law to find the new volume)
First, we need to convert the initial temperature from Celsius to Kelvin:
T1 = 21°C + 273.15 = 294.15 K
Now convert the final temperature from Celsius to Kelvin:
T2 = 175°C + 273.15 = 448.15 K
Now let's apply Charles' Law (V1/T1 = V2/T2):
\(\frac{5.12 \,\text{L}}{294.15 \,\text{K}} = \frac{V_2}{448.15 \,\text{K}}\)
Solving for \(V_2\), we get:
\(V_2 = \frac{5.12 \,\text{L} \cdot 448.15 \,\text{K}}{294.15 \,\text{K}} \approx 7.84 \,\text{L}\)
So, the volume of the gas will be approximately 7.84 L when the temperature is increased to 175°C while the pressure is held constant.
Key Concepts
Boyle's LawCharles' LawIdeal Gas LawTemperature and Pressure Conversion
Boyle's Law
Boyle's Law helps us understand how gases behave when there's a change in pressure, with the temperature constant. This law states that the volume of a gas is inversely proportional to its pressure. In simple terms, if the pressure on a gas increases, its volume decreases, and vice versa. Mathematically, it is represented as:\[ P_1V_1 = P_2V_2 \]Where:
- \( P_1 \) and \( P_2 \) are the initial and final pressures respectively.
- \( V_1 \) and \( V_2 \) are the initial and final volumes respectively.
Charles' Law
Charles' Law explores how gas volume changes with temperature, keeping pressure constant. This law states that the volume of a gas is directly proportional to its temperature, measured in Kelvin. So, if the temperature of a gas increases, its volume increases, provided the pressure remains constant. The equation is:\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]Where:
- \( V_1 \) and \( V_2 \) are the initial and final volumes.
- \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin.
Ideal Gas Law
The Ideal Gas Law is a more comprehensive equation that relates pressure, volume, temperature, and the number of moles of a gas. Though it wasn't directly used in this exercise, understanding it can give a complete picture of gas behaviors. The formula is:\[ PV = nRT \]Where:
- \( P \) is the pressure in atm.
- \( V \) is the volume in liters.
- \( n \) is the number of moles.
- \( R \) is the ideal gas constant \(0.0821 \, ext{L} \, ext{atm} \, ext{mol}^{-1} \, ext{K}^{-1}\).
- \( T \) is the temperature in Kelvin.
Temperature and Pressure Conversion
Temperature and pressure conversions are essential for applying gas laws. Gas law calculations often require temperature to be in Kelvin and pressure in atmospheres. Kelvin (K) is the absolute temperature scale used worldwide for scientific purposes.
- To convert Celsius to Kelvin: add 273.15 (e.g., \(21^{\circ} \, C + 273.15 = 294.15 \, K\)).
- To convert pressure from torr to atm: divide by 760 (e.g., \(752 \, \text{torr} = 0.990 \, \text{atm}\)).
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