Gas Pressure Problem: Step-by-Step Solution

Hey guys! Let's dive into a fascinating problem involving gas behavior. We're going to explore how pressure changes when we mess with the amount of gas and temperature, all while keeping the volume constant. This is a classic physics scenario, and by the end of this article, you'll be a pro at tackling similar problems. So, buckle up and let's get started!

The Initial Scenario: Setting the Stage

Initially, we have a 20 g sample of a gas chilling in a 10.5 L container. The temperature is sitting at 50°C, and the pressure inside is a hefty 1500 mmHg. Think of it like a closed box filled with tiny gas particles bouncing around, creating pressure against the walls. Our mission is to figure out what happens to this pressure when we add more gas and crank up the heat. Remember, the ideal gas law is our trusty tool for solving these types of problems. This law beautifully connects pressure (P), volume (V), the number of moles of gas (n), the ideal gas constant (R), and temperature (T). It's expressed as PV = nRT. In this initial state, we have all the pieces except for the number of moles (n). We can actually figure that out later if needed, but for this problem, we'll focus on how the pressure changes relative to the changes in the amount of gas and temperature.

To truly grasp the situation, let's break down each component. The volume, 10.5 L, is fixed – our container isn't changing size. The temperature, initially at 50°C, plays a crucial role because gas behavior is highly temperature-dependent. We'll need to convert this to Kelvin, the absolute temperature scale, by adding 273.15. So, 50°C becomes 323.15 K. The pressure, 1500 mmHg, gives us a sense of how forcefully the gas particles are colliding with the container walls. And the 20 g of gas? That's our starting mass, which will increase when we add more gas. The ideal gas constant, R, is provided as 62.4 mmHgL/(molK). This constant is a bridge, linking the macroscopic properties of the gas (pressure, volume, temperature) to the microscopic (number of moles).

Imagine the gas molecules as tiny energetic balls bouncing around inside the container. The more balls you have, and the faster they move (higher temperature), the more collisions they'll have with the walls, leading to higher pressure. This intuitive understanding is key to predicting how pressure will change when we alter the conditions. Now, let's move on to the exciting part: adding more gas and heating things up!

The Change: Adding Gas and Increasing Temperature

Now, things get interesting! We're adding 5 g of the same gas to our container, bringing the total gas mass to 25 g. This is like inviting more friends to the party inside the box – more particles are bouncing around. We're also turning up the heat, increasing the temperature to 124°C. Think of this as giving those gas particles an extra energy boost, making them move even faster and collide with more force. The volume, remember, stays constant at 10.5 L. This is crucial because it simplifies our calculations – we only need to consider the changes in pressure due to the changing amount of gas and temperature.

The key here is to understand how the number of moles of gas changes. We added 5 g to the initial 20 g, so we now have 25 g. The number of moles is directly proportional to the mass of the gas. If we double the mass, we double the number of moles (assuming we're dealing with the same gas). Temperature also plays a direct role in pressure, according to the ideal gas law. When we increase the temperature from 50°C to 124°C, we're making the gas particles move significantly faster. This translates to more forceful and frequent collisions with the container walls, which increases the pressure.

To really nail this down, we need to convert the new temperature to Kelvin. Adding 273.15 to 124°C gives us 397.15 K. That's a substantial temperature jump! Now, we have all the pieces we need to calculate the new pressure. We know the initial pressure, the change in the amount of gas, and the change in temperature. We're ready to use the ideal gas law to connect these changes and find our answer. Before we jump into the calculations, let's take a step back and think qualitatively. We've added more gas (more moles) and increased the temperature. Both of these changes should push the pressure upwards. So, we're expecting a final pressure that's higher than our initial 1500 mmHg. This kind of intuitive check is always a good idea to make sure our final answer makes sense.

Calculating the New Pressure: Putting the Pieces Together

Alright, let's get down to the nitty-gritty and calculate the new pressure. This is where the ideal gas law really shines. We're going to use a clever trick to simplify the calculation. Since the volume (V) and the ideal gas constant (R) are constant, we can set up a ratio that relates the initial and final states of the gas. Let's call the initial pressure P1, the initial number of moles n1, and the initial temperature T1. Similarly, the final pressure is P2, the final number of moles is n2, and the final temperature is T2.

From the ideal gas law, we know that P1V = n1RT1 and P2V = n2RT2. Because V and R are constant, we can divide the second equation by the first to get: (P2V) / (P1V) = (n2RT2) / (n1RT1). Notice how the V and R terms cancel out, leaving us with a much simpler equation: P2 / P1 = (n2T2) / (n1T1). This is our golden ticket! It directly relates the initial and final pressures to the changes in the number of moles and temperature.

Now, let's plug in the values we know. We have P1 = 1500 mmHg. The number of moles is proportional to the mass, so we can say n1 is proportional to 20 g and n2 is proportional to 25 g. We have T1 = 323.15 K and T2 = 397.15 K. Plugging these into our equation, we get: P2 / 1500 mmHg = (25 g * 397.15 K) / (20 g * 323.15 K). Notice that we can use the mass directly in this ratio because we're dealing with the same gas. The molar mass, which would be needed to convert grams to moles, cancels out in the ratio.

Now it's just a matter of crunching the numbers. The right side of the equation simplifies to approximately 1.536. So, P2 / 1500 mmHg = 1.536. Multiplying both sides by 1500 mmHg gives us P2 = 2304 mmHg. That's our final answer! The new pressure inside the container is 2304 mmHg. This makes sense – it's higher than the initial pressure, as we predicted.

Final Answer and Key Takeaways

So, after adding 5 g of the gas and increasing the temperature to 124°C, the new pressure inside the 10.5 L container is 2304 mmHg. Awesome! We've successfully navigated this gas pressure problem. But more importantly, we've learned some crucial concepts along the way.

The big takeaway here is the power of the ideal gas law. It's a fundamental equation that describes the behavior of gases and allows us to predict how pressure, volume, temperature, and the amount of gas are related. We saw how keeping the volume constant simplified the problem, allowing us to focus on the changes in the number of moles and temperature. We also learned the importance of converting temperatures to Kelvin – always remember to use the absolute temperature scale when working with gas laws!

Another key point is the ratio method we used. By setting up a ratio of the initial and final states, we eliminated the need to calculate the actual number of moles. This is a handy trick that can save you time and effort in similar problems. Finally, remember the importance of thinking qualitatively about the problem before you start crunching numbers. Predicting the direction of the change (will the pressure increase or decrease?) helps you check if your final answer makes sense.

Gas behavior might seem a bit abstract at first, but it's actually quite intuitive. Think about the gas molecules bouncing around, and how adding more molecules or increasing their speed (temperature) will affect the pressure. With practice, you'll become a gas law guru in no time! Keep practicing, and you'll be able to tackle any gas pressure problem that comes your way.