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A sample of a gas in a 4 L container at 300 K has a pressure of 1 atm. If the temperature is increased to 600 K and the volume is decreased to 2 L, the pressure is likely to be closest to which of the following? A B C D 0.5 atm 1 atm 2 atm 4 atm

A sample of a gas in a 4 L container at 300 K has a pressure of 1 atm. If the temperature is increased to 600 K and the volume is decreased to 2 L, the pressure is likely to be closest to which of the following? A B C D 0.5 atm 1 atm 2 atm 4 atm

Chemistry: Matter and Change
Chemistry: Matter and Change
1st Edition
ISBN:9780078746376
Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Chapter12: States Of Matter
Section12.1: Gases
Problem 12SSC
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**Question:** A sample of a gas in a 4 L container at 300 K has a pressure of 1 atm. If the temperature is increased to 600 K and the volume is decreased to 2 L, the pressure is likely to be closest to which of the following? **Options:** - A. 0.5 atm - B. 1 atm - C. 2 atm - D. 4 atm (correct answer) **Explanation:** Using the ideal gas law and combined gas equation \((P_1 V_1) / T_1 = (P_2 V_2) / T_2\), solve for the final pressure. Initial conditions: \(P_1 = 1\) atm, \(V_1 = 4\) L, \(T_1 = 300\) K. Final conditions: \(V_2 = 2\) L, \(T_2 = 600\) K. Plug the values into the equation: \((1 \text{ atm} \times 4 \text{ L}) / 300 \text{ K} = (P_2 \times 2 \text{ L}) / 600 \text{ K}\) Solving for \(P_2\), the pressure is calculated to be 4 atm.
Transcribed Image Text:**Question:** A sample of a gas in a 4 L container at 300 K has a pressure of 1 atm. If the temperature is increased to 600 K and the volume is decreased to 2 L, the pressure is likely to be closest to which of the following? **Options:** - A. 0.5 atm - B. 1 atm - C. 2 atm - D. 4 atm (correct answer) **Explanation:** Using the ideal gas law and combined gas equation \((P_1 V_1) / T_1 = (P_2 V_2) / T_2\), solve for the final pressure. Initial conditions: \(P_1 = 1\) atm, \(V_1 = 4\) L, \(T_1 = 300\) K. Final conditions: \(V_2 = 2\) L, \(T_2 = 600\) K. Plug the values into the equation: \((1 \text{ atm} \times 4 \text{ L}) / 300 \text{ K} = (P_2 \times 2 \text{ L}) / 600 \text{ K}\) Solving for \(P_2\), the pressure is calculated to be 4 atm.
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