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Chapter 6: Problem 23

A sample of gas has a volume of \(4.25 \mathrm{L}\) at \(25.6^{\circ} \mathrm{C}\) and \(748 \mathrm{mmHg} .\) What will be the volume of this gas at \(26.8^{\circ} \mathrm{C}\) and \(742 \mathrm{mmHg} ?\)

Short Answer

Expert verified
The volume of this gas at \(26.8^{\circ}C\) and \(742 mmHg\) is approximately \(4.28 L\).

Step by step solution

01

Convert Celsius temperature to Kelvin

Gas laws use absolute temperatures, which are measured in Kelvin. The formula to convert from Celsius to Kelvin is \(K = C + 273.15\). Therefore, the initial temperature \(T1 = 25.6^{\circ}C + 273.15 = 298.75 K\) and the final temperature \(T2 = 26.8^{\circ}C + 273.15 = 299.95 K\).
02

Substituting known values into the combined law formula

The combined gas law, which is the volume and temperature proportionality from Charles's Law and the inverse volume and pressure proportionality from Boyle's Law, is expressed as \((P1 x V1) / T1 = (P2 x V2) / T2\). Substituting in the given and calculated values, we get \((748 mmHg x 4.25 L) / 298.75 K = (742 mmHg x V2) / 299.95 K\).
03

Solving for the target variable, V2

The step is solving for final volume \(V2\). This can be done by isolating \(V2\) on one side of the equation by cross multiplication. So, \(V2 = ((748 mmHg x 4.25 L) / 298.75 K) x 299.95 K / 742 mmHg\). Calculating this gives approximately \(4.28 L\).

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Most popular questions from this chapter

Under which conditions is \(\mathrm{Cl}_{2}\) most likely to behave like an ideal gas? Explain. (a) \(100^{\circ} \mathrm{C}\) and \(10.0 \mathrm{atm}\) (b) \(0^{\circ} \mathrm{C}\) and 0.50 atm; \((\mathrm{c}) 200^{\circ} \mathrm{C}\) and \(0.50 \mathrm{atm}\) (d) \(400^{\circ} \mathrm{C}\) and \(10.0 \mathrm{atm}\).

Explain why it is necessary to include the density of \(\mathrm{Hg}(1)\) and the value of the acceleration due to gravity, \(g,\) in a precise definition of a millimeter of mercury (page 194 ).

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