Question

charless law states that the volume of a gas is directly related to the absolute temperature when there is no change in the pressure or amount of gas: (figure 1)
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part b
what celsius temperature, ( t_2 ), is required to change the volume of the gas sample in part a (( t_1 = 45 ^circ \text{c} ), ( v_1 = 1.18 \times 10^3 \text{ l} )) to a volume of ( 2.36 \times 10^3 \text{ l} )? assume no change in pressure or the amount of gas in the balloon.
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review | constants | periodic table

Explanation:

Step1: Recall Charles's Law

Charles's Law is \(\frac{V_1}{T_1}=\frac{V_2}{T_2}\), where temperatures must be in Kelvin. First, convert \(T_1\) from Celsius to Kelvin. \(T_1 = 45^\circ\text{C}+ 273.15=318.15\,\text{K}\), \(V_1 = 1.18\times 10^{3}\,\text{L}\), \(V_2=2.36\times 10^{3}\,\text{L}\).

Step2: Solve for \(T_2\)

Rearrange Charles's Law: \(T_2=\frac{V_2\times T_1}{V_1}\). Substitute values: \(T_2=\frac{2.36\times 10^{3}\,\text{L}\times318.15\,\text{K}}{1.18\times 10^{3}\,\text{L}}\). The \(10^{3}\,\text{L}\) cancels, so \(T_2=\frac{2.36\times318.15}{1.18}\,\text{K}\). Calculate \(2.36\div1.18 = 2\), so \(T_2 = 2\times318.15=636.3\,\text{K}\). Convert back to Celsius: \(T_2=636.3 - 273.15 = 363.15^\circ\text{C}\approx363^\circ\text{C}\) (or keep as Kelvin, but the question asks for Celsius, so convert).

Answer:

The value of \(T_2\) is \(\boldsymbol{363^\circ\text{C}}\) (or \(636\,\text{K}\), but in Celsius as required, approximately \(363^\circ\text{C}\))