Questions: Given R=0.0821 L-atm/mol·K, what is the volume of mole of gas at a pressure of 3.12 atm and a temperature of 270 K? (1 point) 69.2 L/mol 7.10 L/mol 86.5 L/mol 22.2 L/mol

Given R=0.0821 L-atm/mol·K, what is the volume of mole of gas at a pressure of 3.12 atm and a temperature of 270 K? (1 point)
69.2 L/mol
7.10 L/mol
86.5 L/mol
22.2 L/mol

Transcript text: Mark as Comp Gases Unit Test Given $R=\frac{0.0821 \mathrm{~L}-\mathrm{atm}}{\mathrm{mol} \cdot \mathrm{K}}$, what is the volume of mole of gas at a pressure of 3.12 atm and a temperature of 270 K ? (1 point) $69.2 \mathrm{~L} / \mathrm{mol}$ $7.10 \mathrm{~L} / \mathrm{mol}$ 86.5 Lmol $22.2 \mathrm{~L} / \mathrm{mol}$

Solution

Solution Steps

Step 1: Identify the Ideal Gas Law

The Ideal Gas Law is given by: \[ PV = nRT \] where:

$ P $ is the pressure,
$ V $ is the volume,
$ n $ is the number of moles,
$ R $ is the ideal gas constant,
$ T $ is the temperature.

Step 2: Substitute the Given Values

We are given:

$ P = 3.12 $ atm,
$ T = 270 $ K,
$ R = 0.0821 \frac{\mathrm{L} \cdot \mathrm{atm}}{\mathrm{mol} \cdot \mathrm{K}} $,
$ n = 1 $ mole (since we are looking for the volume per mole).

Substitute these values into the Ideal Gas Law: \[ V = \frac{nRT}{P} \]

Step 3: Calculate the Volume

\[ V = \frac{(1 \, \text{mol}) \times (0.0821 \, \frac{\mathrm{L} \cdot \mathrm{atm}}{\mathrm{mol} \cdot \mathrm{K}}) \times (270 \, \text{K})}{3.12 \, \text{atm}} \]

Step 4: Simplify the Expression

\[ V = \frac{0.0821 \times 270}{3.12} \] \[ V = \frac{22.167}{3.12} \] \[ V \approx 7.1058 \, \text{L/mol} \]