Question

2.
Figure shows the variation in internal energy U of 3
moles of Hydrogen gas with volume. The work done
by gas from state A to B is
UA
20
B
U.
EC
D
V.
2V. V
70o in (2)
(1
)
(2) 3 YIN (2)
5
(3) U, In(2)
(4) UV
00​

Answer 1

Explanation:

temp const as internal energy constant

Answer 2

The work done by the gas in the process from A to B is $\frac{4}{5} U_{0} ln2$.

The given process through the points ABCD is a closed cyclic process. Thus, By theory, the total change in internal energy of the system is 0.

                  $delU = 0$

The relation between the Energy, Internal energy, and the work done is :

                 $del Q = delU + delW\\del Q = 0 + delW\\del Q = delW$

Here the total Work will be the sum of the work of all the individual processed AB, BC, CD, and DA.

Work done from states A to B :

• In the process from A to B, the volume changes from V initially to 2V.
• The internal energy in the process remains constant i.e. 2U
• Thus, it is an Isothermal process. Work in this process can be calculated as :

                      $W_{AB} = nRTln(V_{2}/V_{1} )\\W_{AB} = 3RTln(V_{2}/V_{1} )\\W_{AB} = 3RTln(2V/V )\\W_{AB} = 3RTln2$

The internal energy U in terms of the product 'RT' can be calculated as :

                 f = degree of freedom ( 5 here for diatomic gas )

                  $U = (f/2)nRT$

For the process of the AB, we have the internal energy as 2U, so the equation becomes :

                 $2U= (5/2) * 3RT\\\\RT = 4U/15$

Thus, replacing 'RT' in the work equation for AB :

                $W_{AB} = 3RTln2\\W_{AB} = 3(4U/15)ln2\\W_{AB} = (4/5)Uln2$

Hence, the work done by the gas from A to B is $(4/5)U_{0} ln2$.

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