Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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**Gas Laws and Their Applications**

A sample of krypton gas occupies a volume of **6.60 L** at **52.0°C** and **1.16 atm**.

If it is desired to **increase** the volume of the gas sample to **8.17 L**, while **decreasing** its pressure to **0.785 atm**, the temperature of the gas sample at the new volume and pressure must be **_____**°C.

---

To solve this, we'll use the combined gas law, which combines Boyle's Law, Charles's Law, and Gay-Lussac's Law:

\[
\frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2}
\]

Where:
- \( P_1 \) = Initial pressure (1.16 atm)
- \( V_1 \) = Initial volume (6.60 L)
- \( T_1 \) = Initial temperature (52.0°C + 273.15 = 325.15 K)
- \( P_2 \) = Final pressure (0.785 atm)
- \( V_2 \) = Final volume (8.17 L)
- \( T_2 \) = Final temperature (in Kelvin)

Rearranging to solve for \( T_2 \):

\[
T_2 = \frac{P_2 \times V_2 \times T_1}{P_1 \times V_1}
\]

Plugging in the values:

\[
T_2 = \frac{0.785 \times 8.17 \times 325.15}{1.16 \times 6.60}
\]

After calculating, convert \( T_2 \) from Kelvin back to Celsius:

\[
^oC = T_2 - 273.15
\]

Fill up the blank with the calculated temperature for educational purposes.
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Transcribed Image Text:**Gas Laws and Their Applications** A sample of krypton gas occupies a volume of **6.60 L** at **52.0°C** and **1.16 atm**. If it is desired to **increase** the volume of the gas sample to **8.17 L**, while **decreasing** its pressure to **0.785 atm**, the temperature of the gas sample at the new volume and pressure must be **_____**°C. --- To solve this, we'll use the combined gas law, which combines Boyle's Law, Charles's Law, and Gay-Lussac's Law: \[ \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} \] Where: - \( P_1 \) = Initial pressure (1.16 atm) - \( V_1 \) = Initial volume (6.60 L) - \( T_1 \) = Initial temperature (52.0°C + 273.15 = 325.15 K) - \( P_2 \) = Final pressure (0.785 atm) - \( V_2 \) = Final volume (8.17 L) - \( T_2 \) = Final temperature (in Kelvin) Rearranging to solve for \( T_2 \): \[ T_2 = \frac{P_2 \times V_2 \times T_1}{P_1 \times V_1} \] Plugging in the values: \[ T_2 = \frac{0.785 \times 8.17 \times 325.15}{1.16 \times 6.60} \] After calculating, convert \( T_2 \) from Kelvin back to Celsius: \[ ^oC = T_2 - 273.15 \] Fill up the blank with the calculated temperature for educational purposes.
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