An ideal gas in a sealed container has an initial volume of 2.45 L. At a constant pressure, it is cooled to 19.00 °C where its volume is 1.75 L. What was initial temperature?

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An ideal gas in a sealed container has an initial volume of 2.45 L. At a constant pressure, it is cooled to 19.00 °C where its volume is 1.75 L. What was initial temperature?

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Answer 1 · 2018-08-06T03:51:05

$Initial Temperature T_{1} = {135.8}^{\circ} C$ $color(brown)("Initial Temperature " T_1 = 135.8^@ C$

Explanation:

For a given mass of an ideal gas at constant pressure, the volume is directly proportional to its absolute temperature , assuming in a closed system.

![https://www.quora.com/A-perfect-gas-at-27-C-is-heated-at-constant-pressure-till-its-volume-is-doubled-What-will-be-its-final-temperature]( useruploads.socratic.org )

$\frac{V_{1}}{V_{2}} = \frac{T_{1}}{T_{2}}, where T is absolute temperature in KELVIN$ $V_1 / V_2 = T_1 / T_2, " where T is absolute temperature in KELVIN"$

![https://study.com/academy/lesson/http://degrees-celsius-definition-conversion-quiz.html](https://useruploads.socratic.org/6qy1MTlPR5KevFUbwe0i_kelvin.png)

$G i v e n : V_{1} = 2.45 L, V_{2} = 1.75 L, T_{2} = 19^{\circ} C = 273 + 19 = 292^{\circ} K, T_{1} = ?$ $"Given : V_1 = 2.45 L, V_2 = 1.75 L, T_2 = 19^@C = 273 + 19 = 292 ^@ K, T_1 = ?$

$\frac{T_{1}}{T_{2}} = \frac{V_{1}}{V_{2}}$ $T_1 / T_2 = V_1 / V_2$

$T_{1} = (\frac{2.45}{1.75}) \cdot 292 = {408.8}^{\circ} K = 408.8 - 273 = {135.8}^{\circ} C$ $T_1 = (2.45 / 1.75) * 292 = 408.8^@ K = 408.8 - 273 = 135.8^@ C$

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An ideal gas in a sealed container has an initial volume of 2.45 L. At a constant pressure, it is cooled to 19.00 °C where its volume is 1.75 L. What was initial temperature?

1 Answer

Explanation:

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