Question

The temperature of \(2.5 \mathrm{~L}\) of a gas initially at \(\mathrm{STP}\) is raised to \(210^{\circ} \mathrm{C}\) at constant volume. Calculate the final pressure of the gas in atmospheres.

Short answer

The final pressure is approximately 1.77 atm.

Step-by-step solution

Understand the Problem

We have a gas at standard temperature and pressure (STP), which means it is initially at 0°C or 273.15 K and 1 atm pressure. We need to find out the final pressure after the temperature is increased to 210°C. The volume is constant at 2.5 L.

Convert Temperatures to Kelvin

The initial temperature at STP is 0°C, which is equivalent to 273.15 K. The final temperature of the gas is given as 210°C. Convert this to Kelvin using the formula: \[ T(K) = T(°C) + 273.15 \]So, the final temperature is \(210 + 273.15 = 483.15\text{ K}\).

Apply the Ideal Gas Law for Constant Volume

Since volume is constant and we need to relate pressures and temperatures, we can use the formula:\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]Where \(P_1 = 1 \text{ atm}\), \(T_1 = 273.15\text{ K}\), \(P_2\) is the final pressure, and \(T_2 = 483.15\text{ K}\).

Solve for Final Pressure

Rearrange the equation to solve for \(P_2\):\[ P_2 = P_1 \times \frac{T_2}{T_1} \]Plug in the known values:\[ P_2 = 1 \times \frac{483.15}{273.15} \]Calculate \(P_2\):\[ P_2 \approx 1.77 \text{ atm} \]

Interpret the Result

The final pressure of the gas is approximately 1.77 atm. This increase in pressure is due to the temperature rise while maintaining a constant volume.

Key concepts

Temperature Conversion

When dealing with temperature in scientific contexts, we often need to convert temperatures from degrees Celsius (\(°C\)) to Kelvin (\(K\)). This is because many physical formulas, such as those dealing with gases, require Kelvin to ensure consistency and accuracy with the absolute scale.

• To convert from Celsius to Kelvin, simply add 273.15 to the Celsius temperature.
• This adjustment accounts for the absolute zero point, which is 0 K, equivalent to \(-273.15\) °C.

In our exercise, the initial temperature at STP (Standard Temperature and Pressure) was 0°C, which is 273.15 K. After heating, the temperature was 210°C, converting to 483.15 K. This conversion is crucial for further calculations, like determining the gas's pressure change with temperature increasing while keeping the volume constant.

Pressure Calculation

To determine how pressure changes in a gas, the Ideal Gas Law can be adapted for scenarios where certain properties remain constant. The law relates pressure (\(P\)), volume (\(V\)), number of moles (\(n\)), ideal gas constant (\(R\)), and temperature (\(T\)): \( PV = nRT \).

• In our problem, the volume is constant, allowing us to focus on how pressure changes with temperature.
• Using the relationship \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\), we determine final pressure based on initial conditions.

The initial conditions, 1 atm at 273.15 K, allow us to solve for the final pressure at the new temperature, 483.15 K, resulting in a final pressure of approximately 1.77 atm. This method highlights how temperature increases lead to pressure increases when volume remains constant.

STP Conditions

STP, or Standard Temperature and Pressure, is a reference point widely used in chemistry and physics to allow for consistent and comparable results. It signifies conditions of 0°C (273.15 K) and 1 atm pressure.

• STP is convenient for making equations easier to solve and results comparable.
• For gaseous substances, STP allows predictions and calculations about behavior under specified changes, like heating or compression.

Knowing that the exercise started at STP simplifies our calculations as we always have a consistent baseline. This means you can reliably use figures like 273.15 K and 1 atm as initial benchmarks, streamlining problem-solving for gas behavior under varying conditions.

Constant Volume

In our exercise, the volume of the gas did not change, which is a typical scenario when applying the Ideal Gas Law to solve for pressure or temperature. This assumption simplifies the application of gas laws in the form typically called Charles's Law, which relates pressure directly to temperature when volume is constant.

• When the volume is constant, the equation \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\) can be used.
• This means that any increase in temperature will result in a proportional increase in pressure, as long as the gas behaves ideally.

In this scenario, understanding that volume was constant allowed a straightforward calculation of how the pressure changed with temperature, confirming the final pressure increased to 1.77 atm due to the temperature hike to 210°C.