If 3.0 L of oxygen gas at \(177^{\circ} \mathrm{C}\) is cooled at constant pressure until the volume becomes \(1.50 \mathrm{L}\), then what is the final temperature?

Short Answer

Expert verified
The final temperature of the gas is \(225.075K\).

Step by step solution

01

Conversion to Kelvin

First, convert the original temperature from Celsius to Kelvin using the relation: \( T(K) = T(°C) + 273.15 \). The original temperature of \(177^{\circ}C\) becomes \(177 + 273.15 = 450.15K \).
02

Apply Charles's Law

Apply Charles's Law, which can be written: \( \frac{V_1}{T_1} = \frac{V_2}{T_2}\). Here, \(V_1 = 3.0L\) (initial volume), \(T_1 = 450.15K\) (initial temperature), and \(V_2 = 1.5L\) (final volume). The only missing variable is \(T_2\) which we are solving for (the final temperature in Kelvin)
03

Solve for the Final Temperature

Rearrange the equation from Step 2 to solve for \(T_2\): \(T_2 = \frac{V_2 × T_1}{V_1}\). Substituting the known values, we find \(T_2 = \frac{1.5L × 450.15K}{3.0L}= 225.075K\). The final temperature is \(225.075K\)

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Most popular questions from this chapter

Convert each pressure to an equivalent pressure (a) 736 mmHg; (b) 0.776 bar; in atmospheres. (c) 892 Torr; (d) 225 kPa.

A gaseous mixture of He and \(\mathrm{O}_{2}\) has a density of \(0.518 \mathrm{g} / \mathrm{L}\) at \(25^{\circ} \mathrm{C}\) and \(721 \mathrm{mm} \mathrm{Hg} .\) What is the mass percent He in the mixture?

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