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Chapter 10: Problem 28

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.

Short Answer

Expert verified
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.

Step by step solution

01

(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.
02

(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.

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