Question

A gas at \(7.75 \times 10^{4}\) Pa and \(17^{\circ}\) C occupies a volume of 850.0 \(cm^{3} .\) At what temperature, in degrees Celsius, would the gas occupy 720.0 \(cm^{3}\) at \(8.10 \times 10^{4} Pa\) ?

Short answer

41.39°C

Step-by-step solution

Convert Initial Temperature to Kelvin

Given initial temperature is 17°C. Convert this to Kelvin using the formula: \( T_{1} = 17 + 273.15 \). Therefore, \( T_{1} = 290.15 \) K.

Apply Combined Gas Law

The combined gas law is \[ \frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}} \] where \( P_{1} \) = initial pressure, \( V_{1} \) = initial volume, \( T_{1} \) = initial temperature, \( P_{2} \) = final pressure, \( V_{2} \) = final volume, and \( T_{2} \) = final temperature. Plug in the known values: \[ \frac{7.75 \times 10^{4} \, \text{Pa} \times 850.0 \, \text{cm}^{3}}{290.15 \, \text{K}} = \frac{8.10 \times 10^{4} \, \text{Pa} \times 720.0 \, \text{cm}^{3}}{T_{2}} \]

Solve for \( T_{2} \)

Isolate \( T_{2} \) on one side of the equation: \[ T_{2} = \frac{8.10 \times 10^{4} \, \text{Pa} \times 720.0 \, \text{cm}^{3} \times 290.15 \, \text{K}}{7.75 \times 10^{4} \, \text{Pa} \times 850.0 \, \text{cm}^{3}} \] Calculate \( T_{2} \): \[ T_{2} = \frac{8.10 \times 720.0 \times 290.15}{7.75 \times 850.0} \approx 314.54 \, \text{K} \]

Convert \( T_{2} \) to Celsius

Convert the final temperature from Kelvin back to Celsius: \( T_{2} \) (in °C) = 314.54 K - 273.15 = 41.39°C.

Key concepts

Gas Physics

Understanding the behavior of gases is essential in many fields, from chemistry to engineering. The study of gases involves examining how they respond to changes in pressure, volume, and temperature. These relationships are described by various gas laws. The Combined Gas Law is particularly useful because it merges three fundamental gas laws: Boyle's Law, Charles's Law, and Gay-Lussac's Law.
Boyle's Law states that the pressure and volume of a gas are inversely proportional when temperature is constant. Charles's Law tells us that the volume and temperature are directly proportional at constant pressure. Gay-Lussac's Law indicates that pressure and temperature are directly proportional at constant volume. The Combined Gas Law equation can be written as:
\(\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\) This equation allows you to calculate the change in one variable (pressure, volume, or temperature) given the changes in two other variables.

Temperature Conversion

Temperature conversion is a vital skill when working with gas laws because temperatures often need to be converted between Celsius and Kelvin. The Kelvin scale is the standard unit of measurement in scientific contexts because it begins at absolute zero, the point where molecular energy is at its minimum.
To convert Celsius to Kelvin, add 273.15 to the Celsius temperature:
\(\text{Temperature in Kelvin} = \text{Temperature in Celsius} + 273.15\) For example, an initial temperature of 17°C is converted to Kelvin as follows:
\(T_1 = 17 + 273.15 = 290.15 \text{K}\)
To convert Kelvin back to Celsius, subtract 273.15 from the Kelvin temperature.
\(\text{Temperature in Celsius} = \text{Temperature in Kelvin} - 273.15\)
So for a calculated Kelvin temperature of 314.54 K, the equivalent in Celsius would be:
\(T_2 = 314.54 - 273.15 = 41.39 \text{°C}\)

Pressure-Volume Relationship

The pressure-volume relationship forms the foundation of understanding gas behavior. Pressure is defined as the force exerted per unit area, while volume is the space that the gas occupies. The unit of pressure used in this exercise is Pascals (Pa), a standard unit in the SI system.
In Boyle's Law, if you keep the temperature constant, an increase in volume means a decrease in pressure and vice versa. In our example, the initial volume (\(V_1\)) is 850.0 cm³, and the initial pressure (\(P_1\)) is \(7.75 \times 10^{4}\) Pa. Then, we are asked to find the temperature at which the gas's volume becomes 720.0 cm³ at a new pressure of \(8.10 \times 10^{4}\) Pa.
The Combined Gas Law helps us relate changes in these parameters:
\(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\)
Plugging values into the Combined Gas Law equation, we can isolate and solve for the final temperature \(T_2\) to understand how the pressure and volume changes influence the temperature of the gas.