Thermodynamics

To determine the amount of heat extracted from the source in a single cycle of the thermodynamic process represented in the p-V diagram, we need to analyze the components of the cycle.

An ideal monoatomic gas undergoes a cyclic process, and we are provided with a p-V diagram. For such cycles, the heat absorbed or released can be evaluated based on the stages within the cycle.

The key stages often involved are:

• Isothermal process (constant temperature).
• Adiabatic process (no heat exchange).
• Isobaric process (constant pressure).
• Isochoric process (constant volume).

For a monoatomic ideal gas, the specific heat capacity at constant volume and constant pressure are given by:

Considering the amount of heat extracted from the source within these processes, we generally focus on analyzing the net work done over one complete cycle. In cyclic processes for an ideal gas, the net heat exchange, Q, over the entire cycle is equal to the work done, W, by the system:

From the options provided, let us calculate the heat absorbed by examining given values and ensuring they cohesively suggest the correct value:

Upon evaluating the expected thermodynamic model for the gas and corresponding efforts, the proper and consistent answer is:

Hence, the amount of heat extracted from the source in a single cycle, in alignment with the given choice options, is .

The correct answer is (A) : Increases by 250 J
As we know,

( law of thermodynamics)

or


So, the internal energy of the gas increases by

The correct answer is (A) : Both statement-I and statement-II are true

for a polytropic process for adiabatic process r = γ
⇒ Statement I is true
In an adiabatic process
ΔU = –ΔW
⇒ If work is done on the gas
⇒ ΔW is negative
⇒ ΔU is positive or temperature increases
⇒ Statement II is true

Solution for Isothermal and Adiabatic Processes:

Part (A): Isothermal Process

For an isothermal process, the equation is:

Substituting :

Solving for :

Part (B): Adiabatic Process

For an adiabatic process, the equation is:

For a monoatomic gas, . Substituting :

Simplify to find :

Conclusion:

The ratio is 4.

To find the work done by the gas during the part CA, we can use the first law of thermodynamics, which states:

where is the heat absorbed by the system, is the change in internal energy, and is the work done by the system.

Given:

• Heat absorbed during AB = 40 J
• No heat during BC, so
• Heat rejected during CA = -60 J (since it is rejected)
• Work done on the gas during BC = 50 J (negative work done by the gas)
• Internal energy at A,

Since the process is cyclic, the change in internal energy over one complete cycle is zero:

Now, let's analyze the individual processes:

• During AB:
• During BC:
• During CA:

For process CA, using the first law:

Since it's a cyclic process:

From conservation over the full cycle:

Simplifying we see:

(since cycle gives us balance)

By substituting,:

But notice:

• Total input work effect against total heat balance confirms work done, deducing complex heat paths

Correcting refined from question understanding: the effective cyclic cross-check reveals step details guidance points (effective ideal correction):

Net cycles observe border relation current realization (as initially authorized).

So, the correct answer, matching check-form steps, is:

30 J

Fig.

The correct answer is 7479

= 3 × 8.31 × 300
= 7479 J

The correct answer is (D) : W₂ < W₁ < W₃

Comparing the area under the PV graph
A₃ > A₁ > A₂
⇒ W₃ > W₁ > W₂
Therefore , W₂ < W₁ < W₃ is the correct solution

The correct answer is (C) : 27 × 10⁴
Total number of moles are


= 12 moles
Using PV = nRT


= 0.27 m³
= 27 × 10⁴ cm³

The correct answer is : 750

The process described is adiabatic, meaning there is no heat exchange. In such processes, the relation between pressure, volume, and temperature is given by:
. We also know
. Thus, . For an adiabatic change, and . Given , substitute to get:

.

Relating pressures and temperatures, .

Simplifying, .

.

The final temperature is 4 times the initial temperature, which falls within the range of 4,4.