Five moles of gas initially at a pressure of 2.00 atm and a volume of 0.300 L has internal energy equal to 91.0 J. In its final state, the gas is at a pressure of 1.50 atm and a volume of 0.800 L, and its internal energy equals 182 J.
For the paths IAF, IBF, and IF in the figure above, do the following.
(a) calculate the work done on the gas.
(b) Calculate the net energy transferred to the gas by heat in the process.

The work done on the gas is W=-76.0J

The net energy transferred to the gas by heat in the process is 91 J

Explanation:

Given data:

The constant volume of the gas is 0.300L

The initial pressure is 2 atm

The final pressure is 1.50 atm

The formula to calculate the work done is given by,

W= -p(Vf-Vi)

Where,

P denotes the pressure

Vf denotes the final volume

Vi denotes the last volume

Therefore work done on the force is zero, due to the change in the volume is zero

Work done:

Given data

The initial volume of the gas is given by Vi - 0.300L

The final volume of the gas is given by Vf - 0.800L

The constant pressure of the gas is 1.50 atm

The formula to calculate the work done is given by,

W= -p(Vf-Vi)

Where,

P denotes the pressure

Vf denotes the final volume

Vi denotes the initial volume

Substitute the values we get

W= -(1.50 atm (1.30 x 10^5 Pa/atm) (0.800L-0.300L) (10^-3 m ^3/1L)

W=-76.0J

The work done on the gas is W=-76.0J

Net energy

The formula to calcualte the Net energy is given by ΔU =  Uf - Ui

ΔU = Uf - Ui

ΔU = 182 -91 =91

The net energy transferred to the gas by heat in the process is 91 J