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,

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

think its the third option