Question

A \(38.0-\mathrm{L}\) gas tank at \(35^{\circ} \mathrm{C}\) has nitrogen at a pressure of 4.65 atm. The contents of the tank are transferred without loss to an evacuated 55.0 -L tank in a cold room where the temperature is \(4^{\circ} \mathrm{C}\). What is the pressure in the tank?

Short answer

Answer: The final pressure of the nitrogen gas in the second tank is 2.99 atm.

Step-by-step solution

Convert temperatures to Kelvin

Before using the combined gas law, we must first convert the given temperatures from Celsius to Kelvin. To do this, add 273.15 to each temperature: \(T_1 = 35^{\circ} \mathrm{C} + 273.15 = 308.15\,\mathrm{K}\) \(T_2 = 4^{\circ} \mathrm{C} + 273.15 = 277.15\,\mathrm{K}\)

Write down the given values

List the given values to use in the combined gas law formula: \(V_1 = 38.0\,\mathrm{L}\) \(T_1 = 308.15\,\mathrm{K}\) \(P_1 = 4.65\,\mathrm{atm}\) \(V_2 = 55.0\,\mathrm{L}\) \(T_2 = 277.15\,\mathrm{K}\)

Use the combined gas law to find the final pressure

Now, use the combined gas law to find the final pressure, \(P_2\): \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\) Plug in the given values: \(\frac{4.65\,\mathrm{atm} \cdot 38.0\,\mathrm{L}}{308.15\,\mathrm{K}} = \frac{P_2 \cdot 55.0\,\mathrm{L}}{277.15\,\mathrm{K}}\)

Solve for the final pressure, \(P_2\)

Solve the equation for \(P_2\): \(P_2 = \frac{4.65\,\mathrm{atm} \cdot 38.0\,\mathrm{L} \cdot 277.15\,\mathrm{K}}{308.15\,\mathrm{K} \cdot 55.0\,\mathrm{L}}\) \(P_2 = 2.99\,\mathrm{atm}\) The final pressure in the second tank is \(2.99\,\mathrm{atm}\).

Key concepts

Gas Law Calculations

Understanding gas law calculations is fundamental to solving problems in chemistry and physics that involve gases. The real magic happens when we interpret the behavior of gases under various conditions of temperature, volume, and pressure. The combined gas law is one such tool that helps us make these calculations by establishing a relationship between these three variables. To use this law effectively, we first need to collect the given values for the initial and final states of the gas.

It's important to ensure that these values are in the right units, which, for temperature, is always Kelvin. Why Kelvin? Because this scale starts at absolute zero, which is vital from a theoretical standpoint where gases would theoretically occupy zero volume. Converting Celsius to Kelvin is simple: add 273.15 to the Celsius temperature. Once we have everything in the right units, we plug the values into the gas law's formula. The secret to making these calculations seamless is to keep your units consistent and to confidently rearrange the formula to solve for your unknown variable.

Gas Temperature Pressure Relationship

The relationship between gas temperature and pressure is another cornerstone of gas behavior and can be described by Gay-Lussac's Law, a principle that specifically says that pressure of a gas is directly proportional to its temperature, assuming the volume is held constant. However, in the combined gas law, we see this relationship become part of a larger dance, where volume can also change.

Let's consider the nifty mnemonic 'directly proportional, inversely awesome!' Here's why: if the temperature of a gas goes up, assuming volume is unchanged, so will the pressure – they're directly proportional. On the flip side, if we now allow the volume to increase (assuming temperature is unchanged), the pressure will drop – they're inversely related. In the combined gas law formula, the temperature and pressure are on opposite sides, which keeps their direct relationship in check. Mastering this interplay is key to navigating through gas law problems and understanding the behavior of gases in practical applications.

Ideal Gas Law Applications

When we step into the realm of ideal gas law applications, we're opening a vast portal to understanding how gases behave in theory, and mostly in practice. The ideal gas law, represented as PV = nRT, is a more encompassing law where n is the number of moles of the gas and R is the universal gas constant.

This law assumes that gases are made of particles that have no volume themselves and do not interact with each other - an 'ideal' scenario. Although no real gas behaves perfectly as an ideal gas, this law gives incredibly accurate approximations for gases at normal temperatures and pressures. The power of the ideal gas law lies in its broad applications, from calculating the molar mass of a gas to determining stoichiometric relationships in chemical reactions. It provides a solid foundation for understanding more complex systems and can even be applied to more extreme conditions by its adjusted versions, like the Van Der Waals equation for non-ideal gases.