Question

What will the volume of the sample become if $459\mathrm{mL}$ of an ideal gas at ${27}^{\circ }\mathrm{C}$ and 1.05 atm is cooled to 15 C and 0.997 atm?

Short answer

The final volume of the sample under the given conditions will be approximately $473.74\mathrm{mL}$.

Step-by-step solution

Convert temperatures to Kelvin

To work with the combined gas law, we need to convert the given temperatures from Celsius to Kelvin. We can do this using the following formula: $T_K=T_C+273.15$ Initial temperature in Kelvin (T1): $T_1={27}^{\circ }\mathrm{C}+273.15=300.15\mathrm{K}$ Final temperature in Kelvin (T2): $T_2={15}^{\circ }\mathrm{C}+273.15=288.15\mathrm{K}$

Plug in given values into the combined gas law

Now that we have the temperature values in Kelvin, we can plug all of the given values into the combined gas law formula: $\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$ Where: - $P_1=1.05\mathrm{atm}$ - $V_1=459\mathrm{mL}$ - $T_1=300.15\mathrm{K}$ - $P_2=0.997\mathrm{atm}$ - $T_2=288.15\mathrm{K}$ - $V_2$ is the final volume we need to find

Rearrange equation and solve for final volume

To find $V_2$, we first need to rearrange the combined gas law formula to isolate $V_2$ on one side: $V_2=\frac{P_1{V}_1{T}_2}{P_2{T}_1}$ Now, we can plug in the given values and solve for the final volume: $V_2=\frac{(1.05\mathrm{atm})(459\mathrm{mL})(288.15\mathrm{K})}{(0.997\mathrm{atm})(300.15\mathrm{K})}$

Calculate the final volume

Calculate the final volume by performing the arithmetic: $V_2=\frac{(1.05)(459)(288.15)}{(0.997)(300.15)}\approx 473.74\mathrm{mL}$ Therefore, the final volume of the sample under the given conditions will be approximately 473.74 mL.

Key concepts

Temperature Conversion

To solve many gas law problems, converting temperatures from Celsius to Kelvin is a necessary first step. The Kelvin scale is the standard unit of measurement for thermodynamic temperature in scientific calculations. This is because the Kelvin scale starts at absolute zero, where molecular motion stops, providing a more natural reference point for mathematical computations.
The conversion method is straightforward: simply add 273.15 to the Celsius temperature.

• For example, to convert 27°C to Kelvin: $$T_K=27+273.15=300.15$$
• Similarly, for 15°C: $$T_K=15+273.15=288.15$$

Ensuring temperatures are in Kelvin when applying gas laws helps maintain uniformity and accuracy in your calculations.

Ideal Gas Law

The ideal gas law is a cornerstone of chemistry for understanding relationships among pressure, volume, temperature, and amount of gas. Specifically, the combined gas law, used in our problem, integrates aspects of the ideal gas law to cover scenarios where the amount of gas remains unchanged.
In the combined form, the law is represented as:$$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$$This equation connects initial and final states of gas, considering

• $P_1$ and $P_2$ as initial and final pressures, respectively
• $V_1$ and $V_2$ as initial and final volumes
• $T_1$ and $T_2$ as initial and final temperatures in Kelvin

Using this formula allows you to predict how one parameter changes when others are altered, assuming constant gas quantity.

Arithmetic Calculations

After rearranging the combined gas law equation to solve for the final volume, perform arithmetic calculations carefully to ensure accuracy in your answer.The rearranged equation becomes:$$V_2=\frac{P_1{V}_1{T}_2}{P_2{T}_1}$$To find $V_2$, substitute the known values into the equation:$$V_2=\frac{(1.05)(459)(288.15)}{(0.997)(300.15)}$$Breaking down the calculation:

• First, multiply the numerator: $1.05\times 459\times 288.15$
• Then, multiply the denominator: $0.997\times 300.15$
• Finally, divide the results of those calculations to find $V_2$

Executing each calculation in sequence will yield the final volume as approximately 473.74 mL. Proper arithmetic ensures the integrity of your solution.