"I object to violence because when it appears to do good, the good
is only temporary; the evil it does is permanent."

– Mahatma Gandhi

Today we
will discuss the multiple choice single answer type questions on
thermodynamics, which appeared in JEE (Main) 2014 and JEE (Main) 2013 question
papers.

(1) One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure.
The process BC is adiabatic. The temperatures at A, B and C are 400 K, 800 K
and 600 K respectively. Choose the correct statement:

(1) The change in internal energy in the whole cyclic process is 250R

(2) The change in internal energy in the process CA is 700R

(3) The change in internal energy in the process AB is – 350R

(4) The change in internal energy in the process BC is – 500R

The change in the internal energy

*Î”U*is given by*Î”U = n C*

_{v}

*Î”T*where

*n*is the number of moles of gas,

*C*

_{v}is the molar specific heat of the gas at constant volume and

*Î”T*is the change in the temperature of the gas.

The correct option is (4) since the change in
internal energy in the process BC is 1×(5R/2) ×(–200) which is equal to – 500R.

[Note that the molar specific heat of diatomic ideal
gas at

*constant volume*is (5/2)R where R is the universal gas constant]
(2) The above P-V diagram represents the thermodynamic cycle of an
engine, operating with an ideal monoatomic gas. The amount of heat extracted
from the source in a single cycle is:

(1) P

_{0}V_{0}
(2) (13/2) P

_{0}V_{0}
(3) (11/2) P

_{0}V_{0}
(4) 4 P

_{0}V_{0}
The upward vertical portion (on the left of the P-V
diagram) and the rightward horizontal portion (on the top of the P-V diagram)
represent the extraction of heat by the gas from the source. In the first case,
heat is extracted (from the source) at constant volume V

_{0}(isochoric process) and in the second case, heat is extracted (from the source) at constant pressure 2P_{0 }(isobaric process).
Heat absorbed (Q, let us say) from the source during
the isochoric process is given by

Q =
C

_{v}n Î”T where C_{v}is the molar specific heat of the gas at constant volume, Î”T is the change in the temperature of the gas and n*is the number of moles of gas in the engine.*
Since PV = nRT where R

*is universal gas constant, we have*
Î”T
= (Î”P)V/nR =

*(2P*_{0 }– P_{0})V_{0}/nR
Heat absorbed from the source during the isochoric
process is therefore given by

Q =
C

_{v}n (2P_{0 }– P_{0})V_{0}/nR =**(3/2)P**since C_{0}V_{0}_{v}= (3/2)R for an ideal monoatomic gas.
Heat absorbed (Q’, let us say) from the source
during the isobaric process is given by

Q’
= C

_{p}n Î”T where C_{p}is the molar specific heat of the gas at constant pressure.
Since PV = nRT and the varying quantities are V and
T, we obtain

Î”T
= P(Î”V)/nR =

*2P*_{0}(2V_{0 }– V_{0})/nR
Heat absorbed from the source during the isobaric
process is therefore given by

Q’
= C

_{p}n×2P_{0}(2V_{0 }– V_{0})/nR
Since C

_{p}= (5/2)R for an ideal monoatomic gas, we have
Q’
=

**5P**_{0}V_{0}_{ }
The total amount of heat extracted from the source
in a single cycle is therefore given by

Q

_{1}= Q + Q’ = (3/2)P_{0}V_{0}+ 5P_{0}V_{0}=**(13/2)P**_{0}V_{0}
You can access all questions on thermodynamics
posted on this site by clicking on the label ‘thermodynamics’ below this post.

Essential points to be remembered for working out
questions in thermodynamics can be seen here.