When you make ice cubes, the entropy of water

does not change
increases
decreases
may either increase or decrease depending on the process used

Solution

Entropy (Δs) is the disorder, it decreases when water changes to ice-cube

ΔS = ^ΔQ⁄_T

We consider a thermodynamic system. If ΔU represents the increase in its internal energy and W the work done by the system, which of the following statements is true?

ΔU = – W in an adiabatic process
ΔU = W in an isothermal process
ΔU = –W in an isothermal process
ΔU = W in an adiabatic process

Solution

From first law of thermodynamics,

ΔH =Δu + w

In adiabatic process ΔH = 0

∴ Δu = – w

In a Carnot engine efficiency is 40% at hot reservoir temperature T.For efficiency 50%, what will be the temperature of hot reservoir?

T
²⁄₃T
⁴⁄₅T
⁶⁄₅T

Solution

Which of the following statements about a thermodynamic process is wrong ?

For an adiabatic process ΔE_int= – W
For a constant volume process ΔE_int= +Q
For a cyclic process ΔE_int = 0
For free expansion of a gas ΔE_int > 0

Solution

For adiabatic process Q = 0.

By first law of thermodynamics,

Q =ΔE + W

⇒ ΔE_int= – W.

A system goes from A to B via two processes I and II as shown in figure. If ΔU₁ and ΔU₂ are the changes in internal energies in the processes I and II respectively, then

relation between ΔU₁ and ΔU₂ can not be determined
ΔU₁ = ΔU₂
ΔU₁ < ΔU₂
ΔU₁ > ΔU₂

Solution

Change in internal energy do not depend upon the path followed by the process. It only depends on initial and final states i.e.,

ΔU₁ = ΔU₂

In the given (V – T) diagram, what is the relation between pressure P₁ and P₂ ?

P₂ > P₁
P₂ < P₁
P₂ = P₁
Cannot be predicted

Solution

A gas is taken through the cycle A→ B→ C→ A, as shown in figure. What is the net work done by the gas ?

1000 J
zero
– 2000 J
2000 J

Solution

W_net= Area of triangle ABC = ¹⁄₂AC × BC

= ¹⁄₂ × 5 × 10^-3 × 4 × 10⁵ = 1000 J

An ideal gas goes from state A to state B via three different processes as indicated in the P-V diagram :

If Q₁, Q₂, Q₃ indicate the heat a absorbed by the gas along the three processes and ΔUQ₁, ΔU₂, ΔU₂ indicate the change in internal energy along the three processes respectively, then

Q₁ > Q₂ > Q₃ and ΔU₁ = ΔU₂ = ΔU₃
Q₃ > Q₂ > Q₁ and ΔU₁= ΔU₂ = ΔU₃
Q₁ = Q₂ = Q₃ and ΔU₁ > ΔU₂ > ΔU₃
Q₃ > Q₂ > Q₁ and ΔU₁ > ΔU₂ > ΔU₃

Solution

Initial and final condition is same for all process

ΔU₁ = ΔU₂= ΔU₃

from first law of thermodynamics

ΔQ= ΔU + ΔW

Work done

ΔW₁ > ΔW₂ > ΔW₃(Area of P.V. graph)

So ΔQ₁ > ΔQ₂ > ΔQ₃

A thermodynamic system is taken through the cycle ABCD as shown in figure. Heat rejected by the gas during the cycle is

2 PV
4 PV
¹⁄₂P V
P V

Solution

∵ Internal energy is the state function.

∵ In cyclie process; ΔU= 0

According to 1st law of thermodynamics

ΔQ = ΔU + W

So heat absorbed

ΔQ= W= Area under the curve

= – (2V) (P) = – 2PV

So heat rejected = 2PV

A mass of diatomic gas(γ= 1.4) at a pressure of 2 atmospheres is compressed adiabatically so that its temperature rises from 27°C to 927°C. The pressure of the gas in final state is

28 atm
68.7 atm
256 atm
8 atm

Solution

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