Question

At what temperature (in \(\mathrm{C}\) ) will a 5.00 -g sample of neon gas exert a pressure of 1.10 atm in a 7.00 -L container?

Short answer

The temperature at which a 5.00-g sample of neon gas will exert a pressure of 1.10 atm in a 7.00-L container is approximately \(105.62^\circ\mathrm{C}\).

Step-by-step solution

Calculate moles of neon gas

Given the mass of the neon gas is \(5.00 \thinspace g\), we can calculate the number of moles using the molar mass, which is \(20.18 \thinspace g/mol\) for neon. \(n = \frac{mass}{molar \thinspace mass}\) \(n = \frac{5.00 \thinspace g}{20.18 \thinspace g/mol} = 0.2477 \thinspace mol\) The number of moles of neon is approximately \(0.2477 \thinspace mol\).

Rearrange the ideal gas law for T

We can rearrange the ideal gas law to solve for T: \(T = \frac{PV}{nR}\) Now plug in the given values for P, V, n, and R.

Calculate the temperature in Kelvin

Substituting the values into the equation: \(T = \frac{(1.10 \thinspace atm)(7.00 \thinspace L)}{(0.2477 \thinspace mol)(0.0821 \frac{L \cdot atm}{mol \cdot K})}\) \(T = \frac{7.70 \thinspace atm \cdot L}{0.02033 \thinspace mol \cdot atm \cdot L/K} = 378.77 \thinspace K\) The temperature is approximately \(378.77 \thinspace K\).

Convert the temperature to Celsius

To convert the temperature from Kelvin to Celsius, subtract 273.15: \(T(^\circ\mathrm{C}) = T(\mathrm{K}) - 273.15\) \(T(^\circ\mathrm{C}) = 378.77 \thinspace K - 273.15 = 105.62^\circ\mathrm{C}\) The temperature at which the neon gas will exert a pressure of \(1.10 \thinspace atm\) in a \(7.00 \thinspace L\) container is approximately \(105.62^\circ\mathrm{C}\).

Key concepts

Moles calculation

When dealing with gases, calculating the number of moles is a crucial first step for using the Ideal Gas Law. Moles measure the amount of substance present. To find the moles of a gas, you start with the mass of the gas and the molar mass.
The molar mass is a fixed property of each element or compound, representing the weight of one mole of that substance. For neon, the molar mass is 20.18 \( \text{g/mol} \).
The formula to calculate the moles is:

• \[ n = \frac{\text{mass}}{\text{molar mass}} \]

Plug in the values for mass and molar mass. For a given mass of 5.00 grams of neon:

• \[ n = \frac{5.00 \text{ g}}{20.18 \text{ g/mol}} = 0.2477 \text{ mol} \]

This tells us that the sample contains approximately 0.2477 moles of neon.
Understanding this step is important because it allows us to use the Ideal Gas Law to find other unknowns like temperature or pressure.

Temperature conversion

In many scientific calculations, we often start with measuring temperatures in the Kelvin scale. This is because Kelvin is the standard SI unit for temperature and forms the baseline for equations in physics and chemistry. Converting Kelvin to Celsius is a simple step but essential to align with common temperature measurements.
The formula you need to convert a temperature from Kelvin to Celsius is:

• \[ T(^{\circ}C) = T(K) - 273.15 \]

Imagine you calculated a temperature of 378.77 Kelvin. To express this in degrees Celsius:

• \[ T(^{\circ}C) = 378.77 \text{ K} - 273.15 = 105.62^{\circ}C \]

Being comfortable with these conversions can greatly help in lab settings or when interpreting scientific data.

Pressure and volume relationship

In the context of gases, the Ideal Gas Law is a great tool to understand how pressure, volume, and temperature interrelate. The formula is:

• \[ PV = nRT \]

Where:

• \( P \) is pressure in atmospheres (atm)
• \( V \) is volume in liters (L)
• \( n \) is number of moles
• \( R \) is the ideal gas constant, approximately 0.0821 \( \frac{L \cdot atm}{mol \cdot K} \)
• \( T \) is temperature in Kelvin (K)

To find the temperature at which a gas exerts a particular pressure, rearrange the equation to solve for \( T \):

• \[ T = \frac{PV}{nR} \]

Consider a scenario where the pressure \( P = 1.10 \) atm, volume \( V = 7.00 \) L, and moles \( n = 0.2477 \) mol:

• \[ T = \frac{(1.10 \cdot 7.00)}{(0.2477 \cdot 0.0821)} = 378.77 \text{ K} \]

Recognizing these relationships also highlights how gases behave and the conditions under which they operate, which is especially important in fields like chemistry and engineering.