Asystem of ideal gas has an initial pressure of 111 kpa and occupies a volume of 5.00 liters. doubling the system’s absolute temperature by means of a constant-pressure process would require an amount of work w. instead, you decide to double the absolute temperature by carrying out two processes in sequence, a constant-pressure process followed by a constant-volume process. in this case, the total work done in the two-process sequence is w/2. calculate the final pressure of the system.

The final pressure is 148 kPa

Solution:

As per the question:

Initial Pressure of the gas, P = 111 kPa

Initial Volume, V = 5.00 l

Now,

In order to calculate the final pressure of the system:

At constant pressure, the work done is 'W'

If the absolute initial temperature be T

When the absolute temperature is doubled, T' =2T

Total work done in the processes, W =

Now,

From the ideal gas eqn:

PV = nRT

PV ∝ T                      (1)

In constant pressure process, work done is given by:

W =                   (2)

Thus from eqn (1) and (2):

In the first process, temperature is doubled:

V = 2V'

Thus

In the process, the change in volume is given by:

Now,

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .