Using the ideal equation ,determine the value R . Given one gram molecule of a gas at S.T.P occupies 22.4litre.

To find the value of the gas constant \( R \) using the ideal gas equation, we start with the equation itself, which can be expressed as \( PV = nRT \). Here, \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is temperature. At standard temperature and pressure (S.T.P.), we know certain values that will help us calculate \( R \).

Understanding Standard Conditions

At S.T.P., the conditions are defined as a temperature of 273.15 K (0 degrees Celsius) and a pressure of 1 atmosphere (atm).

Given Values

• Volume of one gram molecule of gas: 22.4 liters
• Pressure (P): 1 atm
• Temperature (T): 273.15 K
• Number of moles (n): 1 mole (since we are talking about one gram molecule)

Applying the Ideal Gas Law

We can rearrange the ideal gas equation to solve for \( R \):

R = \frac{PV}{nT}

Substituting the Known Values

Now, we can plug in the known values into the equation:

R = \frac{(1 \text{ atm}) \times (22.4 \text{ L})}{(1 \text{ mole}) \times (273.15 \text{ K})}

Calculating \( R \)

Let’s perform the calculation step by step:

• Numerator: \( 1 \text{ atm} \times 22.4 \text{ L} = 22.4 \text{ atm.L} \)
• Denominator: \( 1 \text{ mole} \times 273.15 \text{ K} = 273.15 \text{ mole.K} \)

Now, dividing the numerator by the denominator:

R = \frac{22.4 \text{ atm.L}}{273.15 \text{ mole.K}} \approx 0.08206 \text{ atm.L/(mol.K)}

Final Result

Thus, the calculated value of the gas constant \( R \) is approximately 0.08206 L·atm/(mol·K). This value is widely used in various gas law calculations and is an essential constant in physical chemistry.

Importance of \( R \)

Knowing the value of \( R \) is crucial as it allows scientists and engineers to perform calculations involving gases under a variety of conditions. It helps in understanding how gases behave when subjected to changes in temperature and pressure.