11. (JEE Main 2022 (Online) 26th July Morning Shift )

A monoatomic gas at pressure $P$ and volume $V$ is suddenly compressed to one eighth of its original volume. The final pressure at constant entropy will be :

(A) P

(B) 8P

(C) 32P

(D) 64P

Correct answer is (C)

$P V^{γ} =$ constant

$\Rightarrow P V^{γ} = (P^{'}) {(\frac{v}{8})}^{γ}$ where $γ = 5 / 3$

$\Rightarrow P^{'} = 32 P$

12. (JEE Main 2022 (Online) 25th July Morning Shift )

A certain amount of gas of volume $V$ at $27^{\circ} C$ temperature and pressure $2 \times 10^{7} {Nm}^{- 2}$ expands isothermally until its volume gets doubled. Later it expands adiabatically until its volume gets redoubled. The final pressure of the gas will be (Use $γ = 1.5)$ :

(A) $3.536 \times 10^{5} Pa$

(B) $3.536 \times 10^{6} Pa$

(C) $1.25 \times 10^{6} Pa$

(D) $1.25 \times 10^{5} Pa$

Correct answer is (B)

JEE Main 2022 (Online) 25th July Morning Shift Physics - Heat and Thermodynamics Question 74 English Explanation

Let AB is isothermal process and BC is adiabatic process then for AB process

P_AV_A = P_BV_B

$\Rightarrow$ P_B = 10⁷ Nm^{$-$ 2}

For process BC

P_BV $_{B}^{r}$ = P_C V $_{C}^{r}$

P_C = 3.536 x $\times$ 10⁶ Pa

13. (JEE Main 2022 (Online) 30th June Morning Shift )

A sample of monoatomic gas is taken at initial pressure of 75 kPa. The volume of the gas is then compressed from 1200 cm³ to 150 cm³ adiabatically. In this process, the value of workdone on the gas will be :

(A) 79 J

(B) 405 J

(C) 4050 J

(D) 9590 J

Correct answer is (B)

For monoatomic gas degree of freedom f = 3 and $γ$ = $\frac{5}{3}$

Here for gas,

Initial pressure (P₁) = 75 kPa

Initial volume (V₁) = 1200 cm³

Final volume (V₂) = 150 cm³

Final pressure (P₂) = ?

For adiabatic process,

$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$

$\Rightarrow 75 \times (1200)^{γ} = P_{2} (150)^{γ}$

$\Rightarrow P_{2} = 75 \times {(\frac{1200}{150})}^{\frac{5}{3}}$

$= 75 \times (8)^{\frac{5}{3}}$

$= 75 \times 32$ kPa

= 2400 kPa

Work done in adiabatic process,

$W = \frac{P_{2} V_{2} - P_{1} V_{1}}{1 - γ}$

$= \frac{2400 \times 10^{3} \times 150 \times 10^{- 6} - 75 \times 10^{3} \times 1200 \times 10^{- 6}}{1 - \frac{5}{3}}$

$= \frac{(2400 \times 150 - 75 \times 1200) \times 10^{- 3}}{- \frac{2}{3}}$

$= \frac{- 810000 \times 10^{- 3}}{2}$

$= - 405$ kJ

14. (JEE Main 2022 (Online) 29th June Evening Shift )

Starting with the same initial conditions, an ideal gas expands from volume V₁ to V₂ in three different ways. The work done by the gas is W₁ if the process is purely isothermal, W₂, if the process is purely adiabatic and W₃ if the process is purely isobaric. Then, choose the correct option

(A) W₁ < W₂ < W₃

(B) W₂ < W₃ < W₁

(C) W₃ < W₁ < W₂

(D) W₂ < W₁ < W₃

Correct answer is (D)

JEE Main 2022 (Online) 29th June Evening Shift Physics - Heat and Thermodynamics Question 108 English Explanation

Comparing the area under the PV graph

A₃ > A₁ > A₂

$\Rightarrow$ W₃ > W₁ > W₂

15. (JEE Main 2022 (Online) 28th June Morning Shift )

Given below are two statements :

Statement I : When $μ$ amount of an ideal gas undergoes adiabatic change from state (P₁, V₁, T₁) to state (P₂, V₂, T₂), then work done is $W = \frac{μ R (T_{2} - T_{1})}{1 - γ}$ , where $γ = \frac{C_{p}}{C_{v}}$ and R = universal gas constant.

Statement II : In the above case, when work is done on the gas, the temperature of the gas would rise.

Choose the correct answer from the options given below :

(A) Both Statement I and Statement II are true.

(B) Both Statement I and Statement II are false.

(C) Statement I is true but Statement II is false.

(D) Statement I is false but Statement II is true.

Correct answer is (A)

$W = \frac{μ R (T_{2} - T_{1})}{1 - r}$ for a polytropic process for adiabatic process r = $γ$

$\Rightarrow$ Statement I is true.

In an adiabatic process

$Δ$ U = $-$ $Δ$ W

$\Rightarrow$ If work is done on the gas

$\Rightarrow$ $Δ$ W is negative

$\Rightarrow$ $Δ$ U is positive or temperature increases

$\Rightarrow$ Statement II is true