16. (JEE Main 2020 (Online) 6th September Morning Slot )

Initially a gas of diatomic molecules is contained in a cylinder of volume V₁ at a pressure P₁ and temperature 250 K. Assuming that 25% of the molecules get dissociated causing a change in number of moles. The pressure of the resulting gas at temperature 2000 K, when contained in a volume 2V₁ is given by P₂ . The ratio $\frac{P_{2}}{P_{1}}$ is ________.

Correct answer is (5)

We know, PV = nRT
$∴$ P₁V₁ = nR (250)

and P₂(2V₁) = $\frac{5 n}{4} R \times (2000)$

By Dividing

$\frac{P_{1}}{2 P_{2}}$ = $\frac{4 \times 250}{5 \times 2000}$

$\Rightarrow$ $\frac{P_{1}}{P_{2}} = \frac{1}{5}$

$\Rightarrow$ $\frac{P_{2}}{P_{1}} = 5$

17. (JEE Main 2020 (Online) 4th September Evening Slot )

The change in the magnitude of the volume of an ideal gas when a small additional pressure $Δ$ P is applied at a constant temperature, is the same as the change when the temperature is reduced by a small quantity $Δ$ T at constant pressure. The initial temperature and pressure of the gas were 300 K and 2 atm. respectively.
If | $Δ$ T| = C| $Δ$ P| then value of C in (K/atm.) is _________.

Correct answer is (150)

We know, $P V = n R T$

$∴$ $P Δ V + V Δ P = 0$ (for constant temp.)

and $P Δ V$ = $n R Δ T$ (for constant pressure)

$Δ T = \frac{P Δ V}{n R}$

$Δ P = - \frac{P Δ V}{V}$ ( $Δ V$ is same in both cases)

$\frac{Δ T}{Δ P} = \frac{P Δ V}{n R} \frac{V}{- P Δ V} = \frac{- V}{n R} = - \frac{T}{P}$

[As PV = nRT

$\Rightarrow$ $\frac{V}{n R} = \frac{T}{P}$ ]

$∴$ $| \frac{Δ T}{Δ P} | = | \frac{- 300}{2} | = 150$

18. (JEE Main 2019 (Online) 10th April Evening Slot )

One mole of ideal gas passes through a process where pressure and volume obey the relation $P = P_{0} [1 - \frac{1}{2} {(\frac{V_{0}}{V})}^{2}]$ . Here P₀ and V₀ are constants. Calculate the change in the temperature of the gas if its volume changes form V₀ to 2V₀

(A) $\frac{3}{4} \frac{P_{0} V_{0}}{R}$

(B) $\frac{1}{2} \frac{P_{0} V_{0}}{R}$

(C) $\frac{5}{4} \frac{P_{0} V_{0}}{R}$

(D) $\frac{1}{4} \frac{P_{0} V_{0}}{R}$

Correct answer is (C)

Given $P = P_{o} {1 - \frac{1}{2} {(\frac{V_{o}}{V})}^{2}};$ ...(i)

As n = 1 mole

$∴$ PV = nRT = RT

$\Rightarrow$ P = $\frac{R T}{V}$ ....(ii)

From (i) and (ii), we get

$\frac{R T}{V} = P_{0} [1 - \frac{1}{2} {(\frac{V_{0}}{V})}^{2}]$

$\Rightarrow$ T = $\frac{V}{R} \times P_{0} [1 - \frac{1}{2} {(\frac{V_{0}}{V})}^{2}]$

Case 1 : when V = V₀

then T_i = $\frac{V_{0}}{R} \times P_{0} [1 - \frac{1}{2} {(\frac{V_{0}}{V_{0}})}^{2}]$

= $\frac{V_{0} P_{0}}{2 R}$

Case 2 : when V = 2V₀

then T_f = $\frac{2 V_{0}}{R} \times P_{0} [1 - \frac{1}{2} {(\frac{V_{0}}{2 V_{0}})}^{2}]$

= $\frac{7}{4} \frac{P_{0} V_{0}}{R}$

Then $Δ$ T = T_f - T_i

= $\frac{7}{4} \frac{P_{0} V_{0}}{R} - \frac{P_{0} V_{0}}{2 R}$

= $\frac{P_{0} V_{0}}{R} (\frac{7}{4} - \frac{1}{2})$

= $\frac{5}{4} \frac{P_{0} V_{0}}{R}$

19. (JEE Main 2019 (Online) 12th January Evening Slot )

A vertical closed cylinder is separated into two parts by a frictionless piston of mass m and of negligible thickness. The piston is free to move along the length of the cylinder. The length of the cylinder above the piston is $ℓ$ ₁, and that below the piston is $ℓ$ ₂, such that $ℓ$ ₁ > $ℓ$ ₂. Each part of the cylinder contains n moles of an ideal gas at equal temperature T. If the piston is stationary, its mass, m, will be given by :
(R is universal gas constant and g is the acceleration due to gravity)

(A) $\frac{n R T}{g} [\frac{ℓ_{1} - ℓ_{2}}{ℓ_{1} ℓ_{2}}]$

(B) $\frac{R T}{g} [\frac{2 ℓ_{1} + ℓ_{2}}{ℓ_{1} ℓ_{2}}]$

(C) $\frac{n R T}{g} [\frac{1}{ℓ_{2}} + \frac{1}{ℓ_{1}}]$

(D) $\frac{R T}{n g} [\frac{ℓ_{1} - 3 ℓ_{2}}{ℓ_{1} ℓ_{2}}]$

Correct answer is (A)

JEE Main 2019 (Online) 12th January Evening Slot Physics - Heat and Thermodynamics Question 249 English Explanation

P₂A = P₁A + mg

$\frac{n R T . A}{A ℓ_{2}}$ = $\frac{n R T . A}{A ℓ_{1}}$ + mg

nRT $(\frac{1}{ℓ_{2}} - \frac{1}{ℓ_{1}})$ = mg

m = $\frac{n R T}{g} (\frac{ℓ_{1} - ℓ_{2}}{ℓ_{1} . ℓ_{2}})$

20. (JEE Main 2018 (Online) 16th April Morning Slot )

One mole of an ideal monoatomic gas is taken along the path ABCA as show in the PV diagram. The maximum temperature attained by the gas along the path BC is given by :

JEE Main 2018 (Online) 16th April Morning Slot Physics - Heat and Thermodynamics Question 285 English

(A) $\frac{25}{16} \frac{P_{o} V_{o}}{R}$

(B) $\frac{25}{8} \frac{P_{o} V_{o}}{R}$

(C) $\frac{25}{4} \frac{P_{o} V_{o}}{R}$

(D) $\frac{5}{8} \frac{P_{o} V_{o}}{R}$

Correct answer is (B)

Equation of line BC,

P = P₀ $-$ $\frac{2 P_{0}}{V_{0}} (V - 2 V_{0})$

As PV = nRT

$∴$ T = $\frac{P V}{n R}$

= $\frac{P_{0} V - \frac{2 P_{0} V_{2}}{V_{0}} + 4 P_{0} V}{1 \times R}$

(As given n = 1 mole gas)

$\Rightarrow$ T = $\frac{P_{0}}{R} [5 V - \frac{2 V^{2}}{V_{0}}]$

For maximum value of T

$\frac{d T}{d V} = 0$

$∴$ 5 $-$ $\frac{4 V}{V_{0}} = 0$

$\Rightarrow$ V = $\frac{5}{4} V_{0}$

$∴$ T_max = $\frac{P_{0}}{R} [5 \times \frac{5 V_{0}}{4} - \frac{2}{V_{0}} \times \frac{25}{16} {V_{0}}^{2}]$

= $\frac{25}{8} \times \frac{P_{0} V_{0}}{R}$

Kinetic Theory of Gases