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What will be the volume at STP of \(\mathrm{O}_{2}\) gas that occupies \(12.0 \mathrm{~L}\) at \(25.0^{\circ} \mathrm{C}\) assuming constant pressure?

Short Answer

Expert verified
The gas at standard temperature and pressure (STP) occupies a volume of approximately 11.0 L.

Step by step solution

01

Convert The Temperature To Kelvin

First, you must convert the given temperature to Kelvin, since the ideal gas law requires the use of absolute temperatures. The equation to convert Celsius to Kelvin is \(K = C + 273.15\). So, \(T_1 = 25.0°C + 273.15 = 298.15 K\) and \(T_2 = 273.15 K \) which is the standard temperature.
02

Apply The Combined Gas Law Formula To Find The Volume At STP

Now that you have both temperatures in Kelvin, you can use the combined gas law formula, which is \(V_1/T_1 = V_2/T_2\). Plug in the given volume for \(V_1\) and solve for \(V_2\), the volume at STP. Thus, \(V_2 = V_1 * T_2 / T_1 = 12.0L * 273.15 K / 298.15 K\).
03

Calculate

Complete the calculation from the previous step to obtain the volume at STP. This gives you the final volume \(V_2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combined Gas Law
Understanding the Combined Gas Law is essential when you need to predict how a gas will behave under different conditions of temperature, volume, and pressure. It's a unification of Boyle's Law, Charles's Law, and Gay-Lussac's Law. The Combined Gas Law formula is \[\begin{equation}\frac{V_1}{T_1} = \frac{V_2}{T_2},\end{equation}\]if the pressure remains constant, which helps us solve problems where a gas' volume and temperature conditions are changing.
Let's use an example for clarity. Imagine you have a balloon filled with air at room temperature. If you take the balloon into a colder environment, its volume will decrease, which you can predict using the Combined Gas Law. By plugging in the initial volume, initial temperature, the final temperature (after converting it to Kelvin), and solving for the final volume, you will know what to expect. This is exactly how we approached the gas volume calculation at STP for an oxygen sample in our exercise.
Temperature Conversion
When working with temperature in gas calculations, converting to the Kelvin scale is crucial because it is the absolute temperature scale used in scientific equations. Converting Celsius to Kelvin is straightforward; you add 273.15 to your Celsius temperature:
\[\begin{equation}K = C + 273.15.\end{equation}\]This is important because gas laws rely on absolute temperature for accuracy. As seen in our exercise, to calculate the volume of gas at STP, converting the given temperature from Celsius to Kelvin allowed us to proceed with the calculation using the Combined Gas Law formula. Temperature conversion is a fundamental step in numerous scientific computations, ensuring that changes in temperature relate explicitly to energy rather than arbitrary degrees on a different scale.
Ideal Gas Law
The Ideal Gas Law is another cornerstone of understanding gas behavior, represented by the equation:\[\begin{equation}P\cdot V = n\cdot R\cdot T,\end{equation}\]where P stands for pressure, V for volume, n for moles of the gas, R for the ideal gas constant, and T for absolute temperature in Kelvin. This law combines several simpler gas laws and assumes that the gas particles are perfect and do not interact with each other.
When conditions are at STP (Standard Temperature and Pressure), the Ideal Gas Law simplifies further, as R and T have set values: R is approximately 0.0821 L·atm/(mol·K), and STP temperature is always 273.15 K. Using this law, chemists can predict how a gas will behave when it is heated, pressurized, or when it expands, which is invaluable in laboratory and industrial applications.
Kelvin Temperature Scale
The Kelvin temperature scale is an absolute scale where 0 Kelvin is equivalent to absolute zero, the theoretically lowest temperature where all thermal motion ceases. Unlike Celsius or Fahrenheit, Kelvin does not use degrees. The importance of the Kelvin scale in scientific study cannot be overstated; it provides a necessary baseline for temperature measurements that is not arbitrary and is directly related to the kinetic energy of particles.
For any gas law calculations, including the Combined Gas Law and the Ideal Gas Law, temperatures must be in Kelvin to maintain consistency and accuracy. In our exercise, converting Celsius to Kelvin was a critical first step, showing just how essential a firm grasp of the Kelvin scale is in chemistry and other physical sciences.

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Most popular questions from this chapter

(a) How is the volume occupied by a gas related to the number of moles of that gas? (b) What other variable must be held constant to show experimentally that this relationship is valid?

What volume of nitrogen is required to react with \(9 \mathrm{~L}\) of hydrogen under the same conditions? $$ 3 \mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$

A total of \(0.400 \mathrm{~L}\) of hydrogen gas is collected over water at \({ }^{18}{ }^{\circ} \mathrm{C}\). The total pressure is 742 torr. If the vapor pressure of water at \(18^{\circ} \mathrm{C}\) is \(15.5\) torr, what is the partial pressure of hydrogen?

For a gas under a given initial set of conditions, calculate the final value for the variable indicated if the other two variables change as described in the following table. $$ \begin{array}{|l|c|c|c|} \hline \text { Initial volume } & 2.50 \mathrm{~L} & 125 \mathrm{~L} & 455 \mathrm{~mL} \\ \hline \text { Initial pressure } & 0.50 \mathrm{~atm} & 0.250 \mathrm{~atm} & 200.0 \text { torr } \\ \hline \text { Initial temperature } & 20.0^{\circ} \mathrm{C} & 25^{\circ} \mathrm{C} & 300 \mathrm{~K} \\ \hline \text { Final volume } & ? & 62.0 \mathrm{~L} & 200.0 \mathrm{~mL} \\ \hline \text { Final pressure } & 760.0 \text { torr } & 100.0 \text { torr } & ? \\ \hline \text { Final temperature } & 0.0^{\circ} \mathrm{C} & ? & 327^{\circ} \mathrm{C} \\ \hline \end{array} $$

Given a fixed quantity of a gas at constant temperature, calculate the new volume the gas would occupy if the pressure were changed as shown in the following table. $$ \begin{array}{|c|c|c|c|} \hline \begin{array}{c} \text { Initial } \\ \text { Volume } \end{array} & \begin{array}{c} \text { Initial } \\ \text { Pressure } \end{array} & \begin{array}{c} \text { Final } \\ \text { Pressure } \end{array} & \begin{array}{c} \text { Final } \\ \text { Volume } \end{array} \\ \hline 2.50 \mathrm{~L} & 2.50 \mathrm{~atm} & 725 \mathrm{~atm} & ? \\ \hline 6.25 \mathrm{~mL} & 825 \text { torr } & 456 \text { torr } & ? \\ \hline 450 \mathrm{~mL} & 50.0 \text { torr } & 30.0 \text { torr } & ? \\ \hline \end{array} $$

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