QuestionJuly 16, 2025

What temperature (in^circ C) did an ideal gas shift to if it was initially at -10.2^circ C at 4.61 atm and 39.0 L and the pressure was changed to 3.0 atm and the volume changed to 40.78 L? Enter a numeric answer with 1 place after the decimal. Answer square

What temperature (in^circ C) did an ideal gas shift to if it was initially at -10.2^circ C at 4.61 atm and 39.0 L and the pressure was changed to 3.0 atm and the volume changed to 40.78 L? Enter a numeric answer with 1 place after the decimal. Answer square
What temperature (in^circ C) did an ideal gas shift to if it was initially at -10.2^circ C at 4.61 atm and 39.0 L and the pressure was changed to 3.0 atm and the volume changed to 40.78 L? Enter a numeric answer with 1 place after the decimal. Answer square

Solution
4.6(159 votes)

Answer

-9.4 Explanation 1. Convert initial temperature to Kelvin T_1 = -10.2 + 273.15 = 262.95 \, K 2. Apply the Combined Gas Law Use \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} to find T_2. 3. Solve for T_2 T_2 = \frac{P_2 V_2 T_1}{P_1 V_1} = \frac{3.0 \times 40.78 \times 262.95}{4.61 \times 39.0} 4. Calculate T_2 T_2 \approx 263.8 \, K 5. Convert T_2 back to Celsius T_2 = 263.8 - 273.15 = -9.4 \, ^{\circ}C

Explanation

1. Convert initial temperature to Kelvin<br /> $T_1 = -10.2 + 273.15 = 262.95 \, K$<br />2. Apply the Combined Gas Law<br /> Use $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$ to find $T_2$.<br />3. Solve for $T_2$<br /> $T_2 = \frac{P_2 V_2 T_1}{P_1 V_1} = \frac{3.0 \times 40.78 \times 262.95}{4.61 \times 39.0}$<br />4. Calculate $T_2$<br /> $T_2 \approx 263.8 \, K$<br />5. Convert $T_2$ back to Celsius<br /> $T_2 = 263.8 - 273.15 = -9.4 \, ^{\circ}C$
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