Question

What volume will 6.00 moles of argon gas occupy at 780.0 mmHg and 30.0 °C? Round pressure to two decimal places.

145 L

14.4 L

0.191 L

0.0189 L

Answer 1

To find the volume that 6.00 moles of argon gas will occupy at a pressure of 780.0 mmHg and a temperature of 30.0 °C, we can use the Ideal Gas Law:

\[ PV = nRT \]

Where:
- \( P \) = pressure in atm
- \( V \) = volume in liters
- \( n \) = number of moles
- \( R \) = ideal gas constant = 0.0821 L·atm/(mol·K)
- \( T \) = temperature in Kelvin

### Step 1: Convert the pressure from mmHg to atm
To convert mmHg to atm, use the conversion factor:
\[ 1 \text{ atm} = 760 \text{ mmHg} \]
So,
\[ P = \frac{780.0 \text{ mmHg}}{760 \text{ mmHg/atm}} \approx 1.0263 \text{ atm} \]

### Step 2: Convert the temperature from °C to K
To convert Celsius to Kelvin, add 273.15:
\[ T = 30.0 + 273.15 = 303.15 \text{ K} \]

### Step 3: Substitute the values into the Ideal Gas Law
Now, we can substitute the known values into the Ideal Gas Law:
- \( n = 6.00 \) moles
- \( R = 0.0821 \text{ L·atm/(mol·K)} \)
- \( T = 303.15 \text{ K} \)
- \( P \approx 1.0263 \text{ atm} \)

Rearranging the Ideal Gas Law to solve for \( V \):
\[ V = \frac{nRT}{P} \]

Now plug in the values:
\[ V = \frac{(6.00 \text{ moles}) (0.0821 \text{ L·atm/(mol·K)}) (303.15 \text{ K})}{1.0263 \text{ atm}} \]

Calculating the numerator:
\[ 6.00 \times 0.0821 \times 303.15 \approx 149.495 \text{ L·atm} \]

Now divide by the pressure:
\[ V \approx \frac{149.495}{1.0263} \approx 145.0 \text{ L} \]

After rounding to two decimal places, we find:
\[ V \approx 145.00 \text{ L} \]

Thus, the final answer is:
**145 L**