Question

A large cylindrical tank contains 0.750 m\(^3\) of nitrogen gas at 27\(^\circ\)C and 7.50 \(\times\) 10\({^3}\)Pa (absolute pressure). The tank has a tight-fitting piston that allows the volume to be changed. What will be the pressure if the volume is decreased to 0.410 m\(^3\) and the temperature is increased to 157\(^\circ\)C?

Short answer

The final pressure is approximately 2.613 x 10^4 Pa.

Step-by-step solution

Understand the Ideal Gas Law

The ideal gas law is given by the equation \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles of gas, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. Since the number of moles and the gas constant remain constant, we can use the relation \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) to find the new pressure \(P_2\).

Convert Temperatures to Kelvin

Convert the given temperatures from Celsius to Kelvin using the formula \(T(\text{K}) = T(\text{°C}) + 273.15\). For the initial temperature \(T_1 = 27\,^ ext{°C}\), so \(T_1 = 27 + 273.15 = 300.15\, \text{K}\). For the final temperature \(T_2 = 157\,^ ext{°C}\), \(T_2 = 157 + 273.15 = 430.15\, \text{K}\).

Use the Pressure-Volume-Temperature Relation

Using the formula \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\), plug in the values: \(P_1 = 7.50 \times 10^3\, \text{Pa}\), \(V_1 = 0.750\, \text{m}^3\), \(T_1 = 300.15\, \text{K}\), \(V_2 = 0.410\, \text{m}^3\), \(T_2 = 430.15\, \text{K}\).

Solve for the Final Pressure, P_2

Rearrange the equation to solve for \(P_2\): \[ P_2 = \frac{P_1V_1T_2}{T_1V_2} \] Insert the known values: \[ P_2 = \frac{(7.50 \times 10^3\, \text{Pa}) \times (0.750\, \text{m}^3) \times (430.15\, \text{K})}{(300.15\, \text{K}) \times (0.410\, \text{m}^3)} \] Calculate \(P_2\) to find the final pressure.

Final Calculation

Perform the calculation for the final pressure: \[ P_2 = \frac{7.50 \times 10^3 \times 0.750 \times 430.15}{300.15 \times 0.410} \approx 2.613 \times 10^4\, \text{Pa} \] This is the new pressure when the volume is reduced and temperature is increased.

Key concepts

Pressure-Volume-Temperature Relation

The relationship between pressure, volume, and temperature of a gas is fundamentally described by the Ideal Gas Law. This law is a key principle in understanding how gases behave under different conditions. The Ideal Gas Law equation is written as \( PV = nRT \) where \( P \) denotes pressure, \( V \) is volume, \( n \) stands for moles of gas, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin. When the number of moles and the gas constant remain constant, changes in pressure, volume, or temperature for a gas sample can be predicted using the equation:\[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \] This specific relation helps us understand how the gas adjusts when the volume changes and temperature varies. Importantly, as temperature increases, the pressure will also increase if the volume decreases, and vice versa. Therefore, the Ideal Gas Law allows us to solve real-world problems involving gases in containers like the example given by calculating how the pressure changes when the physical conditions are altered.

Temperature Conversion to Kelvin

Temperature is a crucial component in the Ideal Gas Law. However, we measure temperature here not in Celsius or Fahrenheit, but in Kelvin. Kelvin is the absolute temperature scale used in scientific calculations because it starts from absolute zero, the point at which there is no kinetic energy in particles. Converting Celsius to Kelvin is straightforward with this formula:

• \( T(\text{K}) = T(\text{°C}) + 273.15 \)

Convert each temperature needed in a calculation. For example, converting an initial temperature of \(27\,°C\) results in:

• \( T_1 = 27 + 273.15 = 300.15\, \text{K} \)

And for a final temperature of \(157\,°C\), it becomes:

• \( T_2 = 157 + 273.15 = 430.15\, \text{K} \)

Only by ensuring all temperatures are in Kelvin can we accurately and reliably use the Ideal Gas Law equations.

Gas Pressure Calculation

Calculating the pressure of a gas residing in a container is a common task that necessitates understanding the interplay of pressure, volume, and temperature. After converting temperatures to Kelvin, you can use the pressure-volume-temperature relation to solve for the new pressure when conditions change.Here's how the calculation typically proceeds:First, gather all known quantities, then apply them in the formula rearranged to solve for \( P_2 \), which is the new pressure:\[ P_2 = \frac{P_1V_1T_2}{T_1V_2} \]By substituting the known values:

• Initial pressure \( P_1 = 7.50 \times 10^3 \text{ Pa} \)
• Initial volume \( V_1 = 0.750 \text{ m}^3 \)
• Initial temperature \( T_1 = 300.15 \text{ K} \)
• Final volume \( V_2 = 0.410 \text{ m}^3 \)
• Final temperature \( T_2 = 430.15 \text{ K} \)

Insert these into the equation:\[ P_2 = \frac{(7.50 \times 10^3 \text{ Pa}) \times (0.750 \text{ m}^3) \times (430.15 \text{ K})}{(300.15 \text{ K}) \times (0.410 \text{ m}^3)} \]Calculate \( P_2 \) to finally find that the new pressure is approximately \( 2.613 \times 10^4 \text{ Pa} \). This process showcases the practical application of the Ideal Gas Law in predicting how a gas's pressure changes in response to shifts in volume and temperature.