QuestionMay 1, 2025

A gas at 49.3^circ C and 893 mm Hg experiences a temperature change and ends up with a pressure of 778 mm Hg. What is the new temperature of the gas in degrees Celsius? T=[?]^circ C

A gas at 49.3^circ C and 893 mm Hg experiences a temperature change and ends up with a pressure of 778 mm Hg. What is the new temperature of the gas in degrees Celsius? T=[?]^circ C
A gas at 49.3^circ C and 893 mm Hg experiences a temperature change and ends up with a pressure of 778 mm Hg. What is the new temperature of the gas in degrees Celsius? T=[?]^circ C

Solution
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Answer

7.62^{\circ}C Explanation 1. Convert Celsius to Kelvin Initial temperature T_1 = 49.3^{\circ}C = 49.3 + 273.15 = 322.45 \, K. 2. Apply the Combined Gas Law Use **\frac{P_1}{T_1} = \frac{P_2}{T_2}**. Given P_1 = 893 \, mm \, Hg, P_2 = 778 \, mm \, Hg, and T_1 = 322.45 \, K. Solve for T_2: T_2 = \frac{P_2 \cdot T_1}{P_1} = \frac{778 \cdot 322.45}{893}. 3. Calculate T_2 T_2 \approx 280.77 \, K. 4. Convert Kelvin to Celsius T_2 = 280.77 - 273.15 = 7.62^{\circ}C.

Explanation

1. Convert Celsius to Kelvin<br /> Initial temperature $T_1 = 49.3^{\circ}C = 49.3 + 273.15 = 322.45 \, K$.<br />2. Apply the Combined Gas Law<br /> Use **$\frac{P_1}{T_1} = \frac{P_2}{T_2}$**. Given $P_1 = 893 \, mm \, Hg$, $P_2 = 778 \, mm \, Hg$, and $T_1 = 322.45 \, K$. Solve for $T_2$: <br /> $T_2 = \frac{P_2 \cdot T_1}{P_1} = \frac{778 \cdot 322.45}{893}$.<br />3. Calculate $T_2$<br /> $T_2 \approx 280.77 \, K$.<br />4. Convert Kelvin to Celsius<br /> $T_2 = 280.77 - 273.15 = 7.62^{\circ}C$.
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