Question

The Volume of 2.8 g CO at 27 degree C and 0.821 atm pressure is (R = 0.08210 Lit. atm \(\text{K}^{-1} \text{mol}^{-1}\))

• A. 3 litres
• B. 0.3 litres
• C. 30 litres
• D. 1.5 litres

Answer 1

The Correct Option is A

Correct Answer: 3 litres

Explanation:
We use the Ideal Gas Equation: 
PV = nRT

Where:
• P = Pressure = 0.821 atm
• V = Volume (what we need to find)
• n = Number of moles of CO
• R = 0.0821 L·atm·K⁻¹·mol⁻¹
• T = Temperature in Kelvin = 27°C + 273 = 300 K

First, calculate the number of moles (n):
Molar mass of CO = 12 (C) + 16 (O) = 28 g/mol
\(n = \frac{2.8}{28} = 0.1 \text{ mol}\)

Now plug into the equation:
\(V = \frac{nRT}{P} = \frac{0.1 \times 0.0821 \times 300}{0.821}\)
\(V = \frac{2.463}{0.821} \approx 3 \text{ litres}\)

So, the volume is 3 litres.

Answer 2

To determine the volume of CO (carbon monoxide) gas at the given conditions, we can use the ideal gas law equation:
\(PV = nRT\)
First, we need to convert the temperature from Celsius to Kelvin:
\(T(K) = T(C) + 273.15\)
\(T(K) = 27 + 273.15 = 300.15 K\)
Next, we can calculate the number of moles of CO:
\(n = \frac{\text{mass}}{\text{molar mass}}\)

\(n = \frac{2.8 \, \text{g}}{12.01 \, \text{g/mol} + 16.00 \, \text{g/mol}}\)

\(n = \frac{2.8 \, \text{g}}{28.01 \, \text{g/mol}}\)
n = 0.099964 mol (approximately 0.1 mol)
Now, we can substitute the values into the ideal gas law equation:
\(PV = nRT\)

\(V = \frac{{nRT}}{{P}}\)

\(V = \frac{{(0.1 \, \text{mol})(0.08210 \, \text{L atm K}^{-1} \, \text{mol}^{-1})(300.15 \, \text{K})}}{{0.821 \, \text{atm}}}\)
V = 3.007 L (approximately 3 L)
Therefore, the volume of 2.8 g of CO at 27 degrees Celsius and 0.821 atm pressure is approximately (A) 3 litres.