JEE Advanced Model Paper 6 with solutions for online practice

A) As the ideal gas expands in vacuum, no work is done (W = 0). Also the container isinsulated therefore no heat is lost or gained (Q = 0). According to first law ofthermodynamicsΔU = Q + W
∴ ΔU = 0 ⇒ There is no change in the temperature of the gas
B) Given PV2 = constant ..... (i)
Also for an ideal gas

PV

T

= constant
From (i) & (ii) V × T = constantAs the gas expands its volume increases and temperature decreases∴ (p) is the correct option
To find whether heat is released or absorbed let us find a relationship between Q and
change in temperature ΔT
We know that Q = nCΔT ..... (i)
Where C = molar specific heat
Also for a polytripic process we have
C = Cv +

R

1−n

and PVn = constant
Here PV2 = Constant. Therefore n =2
∴ C = Cv +

R

1−2

= Cv - R
For mono atomic gas
Cv =

3

2

R
∴ C =

3

2

R - R =

R

2

Substituting this value in (1) we get
Q = n ×

R

2

× ΔT
In this case the temperature decreases i.e. ΔT is negative. Therefore Q is negative. Thisin turn means that heat is lost by the gas during the process. (r ) is the correct option.
C) Proceeding in the same way we get in this case V1∕3 × T = constant
⇒ As the gas expands and volume increases, the temperature decreases. Therefore (p) isthe correct option
In this process, x =

4

3

∴ C = Cv +

R

1− 4 3

=

3

2

R +

3R

−1

=

3

2

R - 3R =

−3R

2

∴ Q = n (

−3R

2

)Δt
As ΔT is negative, Q is positive. This in turn means that heat is gained by the gas duringthe process (s) is the correct option
D) Also ΔT =

Δ(PV)

nR

Here Δ(PV) is posiitve ∴ ΔT is posiitve ∴ temperature increase increases(q) is thecorrect option From the graph it is clear that during the process the pressure of the gasincreases which shows that the internal energy of the gas has increased. Also the volumeincreases which means work is done by the system which needs energy. From these twointerpretation we can comfortably conclude that the gas gains heat during the process.(s) is the correct option.