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1. ⇒ (NEET 2023 Manipur )

For the given cycle, the work done during isobaric process is:

NEET 2023 Manipur Physics - Heat and Thermodynamics Question 1 English

(A) 200 J

(B) Zero

(C) 400 J

(D) 600 J

Correct answer is (D)

NEET 2023 Manipur Physics - Heat and Thermodynamics Question 1 English Explanation

$AB$ is isobaric process

$\begin{aligned} W_{AB} = P (V_{2} - V_{1}) \\ W_{AB} = 3 \times 10^{2} (3 - 1) \\ W_{AB} = 3 \times 100 \times 2 \\ W_{AB} = 600 J \end{aligned}$

2. ⇒ (NEET 2022 Phase 2 )

An ideal gas follows a process described by the equation $P V^{2} = C$ from the initial $(P_{1}, V_{1}, T_{1})$ to final $(P_{2}, V_{2}, T_{2})$ thermodynamic states, where C is a constant. Then

(A) If $P_{1} > P_{2}$ then $V_{1} > V_{2}$

(B) If $P_{1} > P_{2}$ then $T_{1} < T_{2}$

(C) If $V_{2} > V_{1}$ then $T_{2} > T_{1}$

(D) If $V_{2} > V_{1}$ then $T_{2} < T_{1}$

Correct answer is (D)

We know,

$P V = n R T$

Given, $P V^{2} =$ constant

$\Rightarrow$ If $P_{1} > P_{2} \Rightarrow V_{1} < V_{2}$

$P {(\frac{n R T}{P})}^{2} =$ constant

$P^{- 1} T^{2} =$ constant

$\Rightarrow P \propto T^{2}$

i.e., if $P_{1} > P_{2} \Rightarrow T_{1} > T_{2}$

Also, $(\frac{n R T}{V}) V^{2} =$ constant

$T V =$ constant

$\Rightarrow$ If $V_{2} > V_{1}$ then $T_{2} < T_{1}$

3. ⇒ (NEET 2022 Phase 1 )

An ideal gas undergoes four different processes from the same initial state as shown in the figure below. Those processes are adiabatic, isothermal, isobaric and isochoric. The curve which represents the adiabatic process among 1, 2, 3 and 4 is

NEET 2022 Phase 1 Physics - Heat and Thermodynamics Question 9 English

(A) 1

(B) 2

(C) 3

(D) 4

Correct answer is (B)

${(\frac{d P}{d V})}_{a d i a b a t i c} = - γ P$

${(\frac{d P}{d V})}_{i s o t h e r m a l} = - P$

${(\frac{d P}{d V})}_{a d i a b a t i c} > {(\frac{d P}{d V})}_{i s o t h e r m a l}$

NEET 2022 Phase 1 Physics - Heat and Thermodynamics Question 9 English Explanation

4. ⇒ (NEET 2020 Phase 1 )

Two cylinders A and B of equal capacity are connected to each other via a stop cock. A contains an ideal gas at standard temperature and pressure. B is completely evacuated. The entire systems is thermally insulated. The stop cock is suddenly opened. The Process is :

(A) adiabatic

(B) isochoric

(C) isobaric

(D) isothermal

Correct answer is (A)

The system is insulated thermally. So no heat exchange takes place. Therefore the process is adiabatic.

5. ⇒ (NEET 2019 )

In which of the following processes, heat is neither absorbed nor released by a system?

(A) adiabatic

(B) isobaric

(C) isochoric

(D) isothermal

Correct answer is (A)

From the characteristic property of adiabatic process, there is no exchange in heat.

6. ⇒ (NEET 2018 )

The volume (V) of a monatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is NEET 2018 Physics - Heat and Thermodynamics Question 19 English

(A) $\frac{2}{5}$

(B) $\frac{2}{3}$

(C) $\frac{1}{3}$

(D) $\frac{2}{7}$

Correct answer is (A)

Gas is monatomic, so C_p = $\frac{5}{2} R$

Given process is isobaric.

$∴$ dQ = nC_pdT

or dQ = n $(\frac{5}{2} R)$ dT

Also, dW = PdV = nRdT ( $∵$ PV = nRT)

$∴$ Required ratio = $\frac{d W}{d Q}$ = $\frac{n R d T}{n (\frac{5}{2} R) d T}$ = $\frac{2}{5}$

7. ⇒ (NEET 2017 )

Thermodynamic processes are indicated in the following diagram.

NEET 2017 Physics - Heat and Thermodynamics Question 74 English
Match the following

Column-1		Column-2
P.	Process I	A.	Adiabatic
Q.	Process II	B.	Isobaric
R.	Process III	C.	Isochoric
S.	Process IV	D.	Isothermal

(A) P $\to$ C, Q $\to$ A, R $\to$ D, S $\to$ B

(B) P $\to$ C, Q $\to$ D, R $\to$ B, S $\to$ A

(C) P $\to$ D, Q $\to$ B, R $\to$ A, S $\to$ C

(D) P $\to$ A, Q $\to$ C, R $\to$ D, S $\to$ B

Correct answer is (A)

Process I volume is constant hence, it is isochoric

In process IV, pressure is constant hence, it is isobaric.

8. ⇒ (NEET 2016 Phase 2 )

One mole of an ideal monatomic gas undergoes a process described by the equation PV³ = constant. The heat capacity of the gas during this process is

(A) $\frac{3}{2}$ R

(B) $\frac{5}{2}$ R

(C) 2R

(D) R

Correct answer is (D)

Process described by the equation,

PV³ = constant

For a polytropic process, PV^$α$ = constant

$C = C_{V} + \frac{R}{1 - α} = \frac{3}{2} R + \frac{R}{1 - 3} = R$

9. ⇒ (NEET 2016 Phase 1 )

A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then

(A) Compressing the gas isothermally or adiabatically will require the same amount of work.

(B) Which of the case (whether compression through isothermal or through adiabatic process) requires more work will depend upon the atomicity of the gas.

(C) Compressing the gas isothermally will require more work to be done.

(D) Compressing the gas through adiabatic process will require more work to be done.

Correct answer is (D)

NEET 2016 Phase 1 Physics - Heat and Thermodynamics Question 68 English Explanation
V₁ = V, V₂ = V/2

On P-V diagram, Area under adiabatic curve > Area under isothermal curve.

So compressing the gas through adiabatic process will require more work to be done.

10. ⇒ (AIPMT 2015 )

An ideal gas is compressed to half its initial volume by means of several processes. Which of the process results in the maximum work done on the gas ?

(A) Isochoric

(B) Isothermal

(C) Adiabatic

(D) Isobaric

Correct answer is (C)

The P-V diagram of an ideal gas compressed from its initial volume $V_{0}$ to $\frac{V_{0}}{2}$ by several processes is shown in the figure.

AIPMT 2015 Physics - Heat and Thermodynamics Question 65 English Explanation

Work done on the gas = Area under P–V curve

As area under the P–V curve is maximum for adiabatic process, so work done on the gas is maximum for adiabatic process.

11. ⇒ (AIPMT 2014 )

A monatomic gas at a pressure P, having a volume V expands isothermally to a volume 2V and then adiabatically to a volume 16V. The final pressure of the gas is (Take $γ$ = 5/3)

(A) 64P

(B) 32P

(C) P/64

(D) 16P

Correct answer is (C)

First, isothermal expansion

$P V = P^{'} (2 V); P^{'} = \frac{P}{2}$

Then, adiabatic expansion

$P^{'} {(2 V)}^{γ} = P_{f} {(16 V)}^{γ}$
(For adiabatic process, PV^$γ$ = constant)

$\frac{P}{2} (2 V) = P_{f} {(16 V)}^{5 / 3}$

$P_{f} = \frac{P}{2} {(\frac{2 V}{16 V})}^{5 / 3} = \frac{P}{2} {(\frac{1}{8})}^{5 / 3}$

$= \frac{P}{2} {(\frac{1}{2^{3}})}^{5 / 3} = \frac{P}{2} (\frac{1}{2^{5}}) = \frac{P}{64}$

⇒Thermodynamics

Topic 02: Thermodynamic Process Part-1