1 mole of an ideal gas is made to expandfrom an initial volume of 2.24 3dm to11.2 3dm at 300 K. Calculate the maximumamount of work that can be produced inthe surroundings by this process

To calculate the maximum amount of work that can be produced in the surroundings by this process, we can use the formula:

Work = -PΔV

Where:

• Work is the amount of work done by the gas on the surroundings
• P is the pressure of the gas
• ΔV is the change in volume of the gas

First, we need to determine the pressure of the gas. Since the gas is an ideal gas, we can use the ideal gas law:

PV = nRT

Where:

• P is the pressure of the gas
• V is the volume of the gas
• n is the number of moles of the gas
• R is the ideal gas constant (8.314 J/(mol·K))
• T is the temperature of the gas

We are given the initial volume (V1 = 2.24 3dm), the final volume (V2 = 11.2 3dm), and the temperature (T = 300 K). We also know that we have 1 mole of gas (n = 1).

Using the ideal gas law, we can rearrange the equation to solve for the pressure:

P = (nRT) / V

P = (1 mol * 8.314 J/(mol·K) * 300 K) / 2.24 3dm

P = 1110.71 J/dm³

Now that we have the pressure, we can calculate the change in volume (ΔV) by subtracting the initial volume from the final volume:

ΔV = V2 - V1

ΔV = 11.2 3dm - 2.24 3dm

ΔV = 8.96 3dm

Finally, we can calculate the maximum amount of work done by the gas on the surroundings:

Work = -PΔV

Work = -(1110.71 J/dm³ * 8.96 3dm)

Work = -9959.97 J

Therefore, the maximum amount of work that can be produced in the surroundings by this process is -9959.97 J.