16. (JEE Main 2022 (Online) 29th July Morning Shift )

The pressure $P_{1}$ and density $d_{1}$ of diatomic gas $(γ = \frac{7}{5})$ changes suddenly to $P_{2} (> P_{1})$ and $d_{2}$ respectively during an adiabatic process. The temperature of the gas increases and becomes ________ times of its initial temperature. (given $\frac{d_{2}}{d_{1}} = 32$ )

Correct answer is (4)

$P_{1} V_{1}^{γ} = P_{2} V_{2}^{2}$

$\frac{P_{1}}{d_{1}^{γ}} = \frac{P_{2}}{d_{2}^{γ}}$

$\frac{d_{1} T_{1}}{d_{1}^{γ}} = \frac{d_{2} T_{2}}{d_{2}^{γ}}$

$T_{2} = {(\frac{d_{2}}{d_{1}})}^{γ - 1} T_{1}$

$= (32)^{\frac{2}{5}} T_{1}$

$T_{2} = 4 T_{1}$

17. (JEE Main 2021 (Online) 27th July Evening Shift )

One mole of an ideal gas is taken through an adiabatic process where the temperature rises from 27 $^{\circ}$ C to 37 $^{\circ}$ C. If the ideal gas is composed of polyatomic molecule that has 4 vibrational modes, which of the following is true? [R = 8.314 J mol^{$-$ 1} k^{$-$ 1}]

(A) work done by the gas is close to 332 J

(B) work done on the gas is close to 582 J

(C) work done by the gas is close to 582 J

(D) work done on the gas is close to 332 J

Correct answer is (B)

Since, each vibrational mode, corresponds to two degrees of freedom, hence, f = 3 (trans.) + 3 (rot.) + 4 $\times$ 2 (vib.) = 14

& $γ = 1 + \frac{2}{f}$

$γ = 1 + \frac{2}{14} = \frac{8}{7}$

$W = \frac{n R Δ T}{γ - 1} = - 582$

As W < 0. work is done on the gas.

18. (JEE Main 2021 (Online) 27th July Morning Shift )

In the reported figure, there is a cyclic process ABCDA on a sample of 1 mol of a diatomic gas. The temperature of the gas during the process A $\to$ B and C $\to$ D are T₁ and T₂ (T₁ > T₂) respectively.

JEE Main 2021 (Online) 27th July Morning Shift Physics - Heat and Thermodynamics Question 134 English
Choose the correct option out of the following for work done if processes BC and DA are adiabatic.

(A) W_AB = W_DC

(B) W_AD = W_BC

(C) W_BC + W_DA > 0

(D) W_AB < W_CD

Correct answer is (B)

Work done in adiabatic process = $\frac{- n R}{γ - 1} (T_{f} - T_{i})$

$∴$ $W_{A D} = \frac{- n R}{γ - 1} (T_{2} - T_{1})$

and $W_{B C} = \frac{- n R}{γ - 1} (T_{2} - T_{1})$

$∴$ $W_{A D} = W_{B C}$

19. (JEE Main 2021 (Online) 25th July Morning Shift )

A monoatomic ideal gas, initially at temperature T₁ is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T₂ by releasing the piston suddenly. If l₁ and l₂ are the lengths of the gas column, before and after the expansion respectively, then the value of $\frac{T_{1}}{T_{2}}$ will be :

(A) ${(\frac{l_{1}}{l_{2}})}^{\frac{2}{3}}$

(B) ${(\frac{l_{2}}{l_{1}})}^{\frac{2}{3}}$

(C) $\frac{l_{2}}{l_{1}}$

(D) $\frac{l_{1}}{l_{2}}$

Correct answer is (B)

PV^r = const.

TV^{r
$-$ 1} = const.

$T (l)^{\frac{5}{3} - 1}$ = const.

$\frac{T_{1}}{T_{2}} = {(\frac{l_{2}}{l_{1}})}^{\frac{2}{3}}$

20. (JEE Main 2021 (Online) 18th March Evening Shift )

For an adiabatic expansion of an ideal gas, the fractional change in its pressure is equal to (where $γ$ is the ratio of specific heats) :

(A) $- \frac{1}{γ} \frac{d V}{V}$

(B) $- γ \frac{V}{d V}$

(C) $- γ \frac{d V}{V}$

(D) $\frac{d V}{V}$

Correct answer is (C)

for adiabatic expansion :

PV^$γ$ = const.

$\Rightarrow$ ln P + $γ$ ln v = const.

$\Rightarrow$ differentiating both sides;

$\frac{d p}{p} + γ \frac{d v}{v} = 0$

$\Rightarrow \frac{d p}{p} = - γ \frac{d v}{V}$