what pressure is exerted by 0.30 mole of CL2 in a 4.00L container at 20 degree C ?

To calculate the pressure exerted by a gas, we can use the ideal gas law equation:

PV = nRT

Where:
- P is the pressure of the gas in units of pressure (e.g., atm)
- V is the volume of the gas in units of volume (e.g., liters)
- n is the number of moles of the gas
- R is the ideal gas constant (0.0821 L·atm/mol·K)
- T is the temperature in units of temperature (e.g., Kelvin)

Let's calculate the pressure using this equation:

Given:
- n (number of moles) = 0.30 moles
- V (volume) = 4.00 L
- T (temperature) = 20 °C = 20 + 273.15 = 293.15 K

Substituting the values into the ideal gas law equation:

P * 4.00 L = 0.30 moles * 0.0821 L·atm/mol·K * 293.15 K

Simplifying the equation:

P * 4.00 = 0.30 * 0.0821 * 293.15

P * 4.00 = 7.91065525

Dividing both sides of the equation by 4.00:

P = 7.91065525 / 4.00

P ≈ 1.9786 atm

Therefore, the pressure exerted by 0.30 moles of Cl2 in a 4.00 L container at 20°C is approximately 1.9786 atm.

Use PV = nRT