Question 1: Thermodynamics & KTG

Question

One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, as shown in the figure. Its pressure at A is P
0
. Choose the correct option(s) from the following:
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l31jr735/386a1b75-0e05-42d5-a07d-f2e4c1e7f2ab/4e850110-d124-11ec-b5bf-9d106fd05fc4/file-1l31jr736.png?format=png)

Accepted Answer

The internal energy of one mole of an ideal gas at temperature T is given by U = 3RT/2. The process AB is isothermal i.e., T
A
= T
B
= T
0
, which makes U
A
= U
B
.
The work done in the isothermal process AB is
W
AB
= nRT ln(V
B
/V
A
)
= nRT
0
ln(4V
0
/V
0
) = p
0
V
0
ln4.
Information regarding p and T at C can not be obtained from the given graph. Unless it is mentioned that line BC passes through origin or not.
Hence, the correct options are (a) and (b).
Note:
If we assume that the line BC pass through the origin. In the process BC, the slope V/T is constant. Thus, BC is an isobaric process (as V/T = nR/p, by the ideal gas equation). Thus,
p
C
= p
B
= RT
B
/ V
B
= RT
0
/(4V
0
) = (RT
0
/V
0
)4 = p
0
/4.
Apply the ideal gas equation for the states A and C to get
${T_C} = {{{p_C}{V_C}} \over {{p_0}{V_0}}}{T_0} = {{({p_0}/4){V_0}} \over {{p_0}{V_0}}}{T_0} = {T_0}/4$.

Answer

Internal energies at A and B are the same.

Answer

Work done by the gas in process AB is P
0
V
0
ln4.

Answer

Pressure at C is P
0
/4.

Answer

Temperature at C is T
0
/4.

Thermodynamics & KTG

One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, as shown in the figure. Its pressure at A is P 0 . Choose the correct option(s) from the following:

Choose the correct answer: