Question

An ideal gas is contained in a cylinder with a volume of \(5.0 \times 10^{2} \mathrm{mL}\) at a temperature of \(30 .^{\circ} \mathrm{C}\) and a pressure of \(710.\) torr. The gas is then compressed to a volume of \(25 \mathrm{mL}\) and the temperature is raised to \(820 .^{\circ} \mathrm{C}\). What is the new pressure of the gas?

Short answer

The new pressure of the gas after compression and heating is approximately \(76.7\: atm\).

Step-by-step solution

Convert given values to appropriate units

In this problem, we are working with volumes in milliliters, pressure in torr, and temperature in Celsius. First, we need to convert them into the standard units to work with the ideal gas law: SI units for volume(liters), pressure(atmospheres), and temperature(Kelvin). Volume unit conversion: Initial volume, \(V_1 = 5.0 \times 10^2 \: mL = 0.500\: L\) Final volume, \(V_2 = 25\: mL = 0.025\: L\) Pressure unit conversion: Initial pressure, \(P_1 = 710.0\: torr\) We know that there are \(101325\: Pa\: (1\: atm)\) in \(760\: torr\). To convert the given pressure to atm, we divide by \(760\). \(P_1 = 710.0\: torr \times \frac{1\: atm}{760\: torr} = 0.9342\: atm\) Temperature unit conversion: Initial temperature, \(T_1 = 30.0\: ^\circ C\) We convert from Celsius to Kelvin by adding \(273.15\). \(T_1 = 30.0 + 273.15 = 303.15\: K\) Final temperature, \(T_2 = 820.0\: ^\circ C\) \(T_2 = 820.0 + 273.15 = 1093.15\: K\)

Apply the combined gas law to find the final pressure

Now that we have converted the given values to appropriate units, we can use the combined gas law formula to find the new pressure: \(P_1V_1/T_1 = P_2V_2/T_2\) Rearrange equation to solve for the final pressure, \(P_2\): \(P_2 = \frac{P_1V_1T_2}{V_2T_1}\) Substitute the known values into the equation: \(P_2 = \frac{(0.9342\: atm)(0.500\: L)(1093.15\: K)}{(0.025\: L)(303.15\: K)}\) Calculate \(P_2\): \(P_2 = 76.674\: atm\) So, the new pressure of the gas after compression and heating is approximately \(76.7\: atm\).

Key concepts

Pressure Conversion

Whenever you work with gases, oftentimes you'll encounter pressure values expressed in various units such as torr, atm, or pascals. A very common necessity is to convert these units for consistency, especially when working with the Ideal Gas Law. The Ideal Gas Law typically uses atmospheres (atm) as the pressure unit.

In the given exercise, the initial pressure is provided in torr. Since 1 atm equals 760 torr, you can easily convert pressure from torr to atm by dividing the torr value by 760. For instance, if you have a pressure of 710 torr, you convert it by doing the following:

- Given: 710 torr - Conversion: \( 710 \ \text{torr} \times \frac{1 \ \text{atm}}{760 \ \text{torr}} = 0.9342 \ \text{atm} \)

This conversion is essential for employing the combined gas law later on with standard units.

Temperature Conversion

In gas law calculations, expressing temperature in Kelvin is crucial, as Kelvin is the absolute temperature scale used in these calculations. The relationship between degrees Celsius and Kelvin is straightforward: you add 273.15 to the Celsius temperature to get the Kelvin temperature.

For example, to convert 30°C to Kelvin, you compute:

- Initial temperature: 30°C - Conversion to Kelvin: \( 30 + 273.15 = 303.15 \ \text{K} \)

Similarly, conversion from 820°C involves the same process:

- Final temperature: 820°C - Conversion to Kelvin: \( 820 + 273.15 = 1093.15 \ \text{K} \)

If you want to use gas laws and ensure your answer is reliable, using Kelvin is non-negotiable. It allows for the equations to correctly describe the physical behavior of gases across variable conditions.

Combined Gas Law

The combined gas law is a powerful tool for predicting the behavior of a gas when it is subject to changes in pressure, volume, and temperature. This law merges Boyle's Law, Charles's Law, and Gay-Lussac's Law into one equation.

Its formula is:- \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \)

This equation implies that the ratio of the product of pressure and volume to temperature is constant for a given amount of gas. To solve for the unknown, such as the final pressure in our original problem, you rearrange the equation accordingly:

- Rearranging: \( P_2 = \frac{P_1V_1T_2}{V_2T_1} \)

By plugging in the given values — initial pressure, initial and final volumes, and temperatures — you can find the final pressure after the gas is compressed and heated in the cylinder. This is a practical application of gas laws that observes real-life behavior of gases.

Unit Conversion

When dealing with gases, unit conversion plays an important role in ensuring that all physical quantities are compatible within the equations being used. For volume, it’s common to convert milliliters (mL) to liters (L), since the gas laws typically require standard SI units.

This process is simple, knowing that 1000 mL equals 1 L:

- Initial volume: 500 mL - Conversion to liters: \( 500 \ \text{mL} = 0.500 \ \text{L} \)

- Final volume: 25 mL - Conversion to liters: \( 25 \ \text{mL} = 0.025 \ \text{L} \)

These conversions, along with pressure and temperature conversions, are essential not only for solving the problem but also for ensuring the reliability and accuracy of the results achieved through the calculations. Always ensure consistency in units to avoid errors in your applications.