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.

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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.
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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