A flexible .556 L container holds air at a temperature of 27.3 degrees C. If the container is cooled so that the new volume of the gas is .214 L, what is the new temperature of the air in Celsius?

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Answer 1 · 2018-08-10T00:31:34

$-157 Celsius$ $"-157 Celsius"$

Here are the values obtained from the question:

Since the question involves changes in volume and temperature, we'll need to use Charles' law:

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

Before plugging in the values, the unit for temperature has to be in Kelvin, therefore we'll need to convert from Celsius to Kelvin:

$T_{1} = 27.3 Celsius + 273.15 = 300.45 K$ $T_1 = "27.3 Celsius" + 273.15 = "300.45 K"$

Solving for $T_{2}$ $T_2$ :

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

$\frac{0.556 L}{300.45 K} = \frac{0.214 L}{T_{2}}$ $("0.556 L")/("300.45 K")=("0.214 L")/(T_2)$

$T_{2} = 0.214 L \times \frac{300.45 K}{0.556 L} = 116 K (3 sig. fig.)$ $T_2="0.214 L"xx("300.45 K")/("0.556 L")="116 K" ("3 sig. fig.")$

Since the question is asking for $T_{2}$ $T_2$ in Celsius, we'll need to convert from Kelvin to Celsius:

$T_{2} = 116 K - 273.15 = -157 Celsius$ $T_2="116 K" - 273.15 ="-157 Celsius"$