A gas sample containing \(1.50\) moles at \(25^{\circ} \mathrm{C}\) exerts a pressure of \(400 .\) torr. Some gas is added to the same container and the temperature is increased to \(50 .^{\circ} \mathrm{C}\). If the pressure increases to \(800 .\) torr, how many moles of gas were added to the container? Assume a constant-volume container.

Short Answer

Expert verified
1.50 moles of gas were added to the container.

Step by step solution

01

Convert temperatures to Kelvin

Convert given temperatures from Celsius to Kelvin by adding 273.15 to both the initial and final temperatures: Initial temperature: \[T_1 = 25^{\circ}\mathrm{C} + 273.15 = 298.15\,\mathrm{K}\] Final temperature: \[T_2 = 50^{\circ}\mathrm{C} + 273.15 = 323.15\,\mathrm{K}\]
02

Convert pressures to atmospheres

Convert given pressures from torr to atmospheres: Initial pressure: \[P_1 = \frac{400\,\mathrm{torr}}{760\,\mathrm{torr/atm}} = 0.5263\,\mathrm{atm}\] Final pressure: \[P_2 = \frac{800\,\mathrm{torr}}{760\,\mathrm{torr/atm}} = 1.0526\,\mathrm{atm}\]
03

Write ideal gas law equations

Since the container has a constant volume, we can write two ideal gas law equations representing the initial and final states: Initial state: \(P_1V = n_1RT_1\) Final state: \(P_2V = n_2RT_2\)
04

Find volume from the initial state

We can find the volume in the container using the initial state values: \(P_1V = n_1RT_1 \Rightarrow V = \frac{n_1RT_1}{P_1}\) Volume: \(V = \frac{(1.50\,\mathrm{mol})(0.0821\,\mathrm{L\, atm/mol\, K})(298.15\,\mathrm{K})}{0.5263\,\mathrm{atm}} = 74.64\,\mathrm{L}\)
05

Find the final number of moles

Now that we have the value for V, we can use the final state equation to find \(n_2\): \(P_2V = n_2RT_2\) \(\Rightarrow n_2 = \frac{P_2V}{RT_2}\) Final number of moles: \(n_2 = \frac{(1.0526\,\mathrm{atm})(74.64\,\mathrm{L})}{(0.0821\,\mathrm{L\,atm/mol\,K})(323.15\,\mathrm{K})} = 3.00\,\mathrm{mol}\)
06

Calculate the additional moles of gas

To find the number of moles of gas added to the container, subtract the initial moles from the final moles: \(\Delta n = n_2 - n_1\) \(\Delta n = 3.00\,\mathrm{mol} - 1.50\,\mathrm{mol} = 1.50\,\mathrm{mol}\) 1.50 moles of gas were added to the container.

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Key Concepts

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

Moles of Gas
The concept of 'moles of gas' is fundamental in understanding chemical amounts in gaseous reactions and gas behavior. A mole is a unit that measures the amount of a substance. In the realm of gases, it allows chemists to correlate the number of particles with a volume the gas will occupy at certain conditions of pressure and temperature. It's particularly useful when applying the Ideal Gas Law, as seen in our exercise.

In the exercise, we are informed about the initial moles of gas, and we have to determine how many additional moles are added by examining changes in pressure and temperature. Remember that standard molar volume at STP (Standard Temperature and Pressure) for an ideal gas is approximately 22.4 liters per mole, but this volume will vary with changes in temperature and pressure, as described by the gas law equations.
Gas Laws
Gas laws are essential to understand how gases interact with changes in temperature, pressure, and volume. The most encompassing of these is the Ideal Gas Law, shown as \( PV = nRT \), where \( P \) stands for pressure, \( V \) for volume, \( n \) for the number of moles, \( R \) for the universal gas constant and \( T \) for temperature in Kelvin. This law assumes that gases behave ideally, meaning gas particles do not attract or repel each other, and occupy negligible space.

In the provided exercise, the Ideal Gas Law allows us to connect the initial and final states of the gas in a constant-volume container to discern the amount of gas added. As a side note, when only temperature and pressure change while volume and moles remain fixed, the Gay-Lussac's Law part of the combined gas law is applied; when only moles and pressure change, it's the direct application of Avogadro's Law.
Temperature Conversion
Temperature conversion is a routine step in gas law problems since the Ideal Gas Law requires temperature to be in Kelvin. To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature. This shift ensures the absence of negative temperatures, aligning with the principle that absolute zero (0 K) is the lowest possible temperature at which particles theoretically have minimum thermal motion.

In the exercise, we convert the given temperatures from Celsius to Kelvin to properly use the Ideal Gas Law. It is crucial that you always perform this conversion before plugging the values into the gas law equations to ensure accuracy in your calculations.
Pressure Conversion
Pressure conversion is often needed because pressure can be measured in several units such as atmospheres (atm), torr, or pascals (Pa). In the context of the Ideal Gas Law, pressure is commonly expressed in atmospheres. To convert torr to atmospheres, divide the pressure in torr by 760, since one atmosphere is equivalent to 760 torr.

In our problem, we convert the initial and final pressures from torr to atmospheres to align with the universal gas constant's units \( R = 0.0821 \text{ L atm/mol K} \). Accurate pressure conversion is vital for solving gas law problems because using inconsistent units can lead to incorrect results.

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