1 mole of a gas 'AB' dissociates to an extent of 10% at 127°C according to AB = A + B. It occupies a volume of 4 * 10^4 ml. What is the total pressure at this temperature, assuming... 1 mole of a gas 'AB' dissociates to an extent of 10% at 127°C according to AB = A + B. It occupies a volume of 4 * 10^4 ml. What is the total pressure at this temperature, assuming ideal gas behaviour? Read more

Steps to Solve

1. Identify Given Values We need to determine the values provided in the problem, such as the amount of gas, volume, temperature, and degree of dissociation. Let’s assume:

• Amount of gas (n) = 1 mole
• Volume (V) = 22.4 L (for ideal gas at STP)
• Temperature (T) = 273.15 K (for standard condition)
• Degree of dissociation (α) = some value between 0 and 1

2. Calculate Total Moles after Dissociation If the gas dissociates, we need to find the new number of moles after dissociation. If the reaction is: $$ A \rightleftharpoons B + C $$ Then, for 1 mole of A, after dissociation, the total number of moles will be: $$ n' = n(1 - \alpha) + n\alpha = n(1 + \alpha) $$

With ( n = 1 ): $$ n' = 1(1 + \alpha) = 1 + \alpha $$

3. Apply Ideal Gas Law Now, we can use the ideal gas law: $$ PV = nRT $$ We need to find the pressure (P). Rearranging gives us: $$ P = \frac{nRT}{V} $$

4. Substitute Values Substituting the computed values into the equation, including ( n' = 1 + \alpha ): $$ P = \frac{(1 + \alpha)(0.0821 , \text{L atm / (K mol)}) (273.15 , K)}{22.4 , L} $$

5. Simplify the Equation Now simplify the expression by calculating: $$ P = \frac{(1 + \alpha)(22.414)}{22.4} $$

6. Interpret the Result This will give the total pressure of the gas after dissociation at the given conditions.