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Chapter 5: Problem 49

If 27.1 g of \(\mathrm{Ar}(g)\) occupies a volume of 4.21 \(\mathrm{L}\) , what volume will 1.29 moles of \(\mathrm{Ne}(g)\) occupy at the same temperature and pressure?

Short Answer

Expert verified
At the same temperature and pressure, 1.29 moles of neon gas will occupy approximately 8.14 L of volume.

Step by step solution

01

Identifying the given values

We are given the mass of Ar (argon gas) = 27.1 g, volume occupied by Ar = 4.21 L, and the moles of Ne (neon gas) = 1.29 moles. We need to find the volume occupied by Ne at the same temperature and pressure.
02

Calculating the moles of Ar

To find the moles of Ar, we'll use the equation: moles = mass/molar mass. The molar mass of Argon is 39.95 g/mol. So, we have: moles of Ar = \( \frac{27.1\, \mathrm{g}}{39.95\, \mathrm{g/mol}} \) = 0.678 moles (approximately)
03

Applying the Ideal Gas Law

According to the Ideal Gas Law, \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Since the temperatures and pressures for both gases are the same, we can write: \( \frac{P \times V_{Ar}}{n_{Ar} \times T} = \frac{P \times V_{Ne}}{n_{Ne} \times T} \)
04

Rearranging the equation to solve for V_neon

We'll rearrange the equation to find the volume occupied by Ne: \( V_{Ne} = V_{Ar} \times \frac{n_{Ne}}{n_{Ar}} \)
05

Substitute the values

Now we'll substitute the given values for volume and moles of Ar and moles of Ne: \( V_{Ne} = (4.21\, \mathrm{L}) \times \frac{1.29\, \mathrm{moles\, Ne}}{0.678\, \mathrm{moles\, Ar}} \)
06

Calculate the volume of Ne

By calculating the above expression, we get the volume occupied by Ne: \( V_{Ne} \approx 8.14\, \mathrm{L} \) So, at the same temperature and pressure, 1.29 moles of neon gas will occupy approximately 8.14 L of volume.

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Most popular questions from this chapter

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