Question

charles’s 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)
part a
a sample of gas in a balloon has an initial temperature of 45. °c and a volume of 1.18×10³ l. if the temperature changes to 69. °c, and there is no change of pressure or amount of gas, what is the new volume, v₂, of the gas?
express your answer with the appropriate units.
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part b
what celsius temperature, t₂, is required to change the volume of the gas sample in part a (t₁ = 45. °c, v₁ = 1.18×10³ l) to a volume of 2.36×10³ l? assume no change in pressure or the amount of gas in the balloon.

Explanation:

Step1: Recall Charles's Law

Charles's Law states that for a gas at constant pressure and amount, \(\frac{V_1}{T_1}=\frac{V_2}{T_2}\), where \(V\) is volume and \(T\) is absolute temperature (in Kelvin). First, convert Celsius temperatures to Kelvin. \(T_1 = 45^\circ\text{C}+ 273.15=318.15\,\text{K}\), \(T_2 = 69^\circ\text{C}+ 273.15 = 342.15\,\text{K}\), \(V_1 = 1.18\times 10^{3}\,\text{L}\).

Step2: Rearrange Charles's Law for \(V_2\)

From \(\frac{V_1}{T_1}=\frac{V_2}{T_2}\), we get \(V_2=\frac{V_1\times T_2}{T_1}\).

Step3: Substitute values

Substitute \(V_1 = 1.18\times 10^{3}\,\text{L}\), \(T_1 = 318.15\,\text{K}\), \(T_2 = 342.15\,\text{K}\) into the formula:
\(V_2=\frac{1.18\times 10^{3}\,\text{L}\times 342.15\,\text{K}}{318.15\,\text{K}}\). Calculate the numerator: \(1.18\times 10^{3}\times 342.15 = 1.18\times342150 = 403737\). Then divide by \(318.15\): \(\frac{403737}{318.15}\approx 1269\,\text{L}\) or \(1.27\times 10^{3}\,\text{L}\) (rounded appropriately).

Answer:

The new volume \(V_2\) is approximately \(\boldsymbol{1.27\times 10^{3}\,\text{L}}\) (or \(1270\,\text{L}\) depending on rounding).