11. (JEE Main 2021 (Online) 31st August Morning Shift )

For an ideal gas the instantaneous change in pressure 'p' with volume 'v' is given by the equation $\frac{d p}{d v} = - a p$ . If p = p₀ at v =0 is the given boundary condition, then the maximum temperature one mole of gas can attain is : (Here R is the gas constant)

(A) $\frac{p_{0}}{a e R}$

(B) $\frac{a p_{0}}{e R}$

(C) infinity

(D) 0 $^{\circ}$ C

Correct answer is (A)

$\int_{p_{0}}^{p} \frac{d p}{P} = - a \int_{0}^{v} d v$

$\ln (\frac{p}{p_{0}}) = - a v$

$p = p_{0} e^{- a v}$

For temperature maximum p-v product should be maximum

$T = \frac{p v}{n R} = \frac{p_{0} v e^{- a v}}{R}$

$\frac{d T}{d v} = 0 \Rightarrow \frac{p_{0}}{R} {e^{- a v} + v e^{- a v} (- a)}$ = 0

$\frac{p_{0} e^{- a v}}{R} {1 - a v} = 0$

$v = \frac{1}{a}, \infty$

$T = \frac{p_{0} 1}{R a e} = \frac{p_{0}}{R a e}$

at v = $\infty$

T = 0

Option (a)

12. (JEE Main 2021 (Online) 27th August Morning Shift )

A balloon carries a total load of 185 kg at normal pressure and temperature of 27 $^{\circ}$ C. What load will the balloon carry on rising to a height at which the barometric pressure is 45 cm of Hg and the temperature is $-$ 7 $^{\circ}$ C. Assuming the volume constant?

(A) 181.46 kg

(B) 214.15 kg

(C) 219.07 kg

(D) 123.54 kg

Correct answer is (D)

P_m = $ρ$ RT

$∴$ $\frac{P_{1}}{P_{2}} = \frac{ρ_{1} T_{1}}{ρ_{1} T_{2}}$

$\frac{ρ_{1}}{ρ_{2}} \Rightarrow \frac{P_{1} T_{2}}{P_{2} T_{1}} = (\frac{76}{45}) \times \frac{266}{300}$

$\frac{ρ_{1}}{ρ_{2}} \Rightarrow \frac{M_{1}}{M_{2}} = \frac{76 \times 266}{45 \times 300}$

$∴$ $M_{2} \Rightarrow \frac{45 \times 300 \times 185}{76 \times 266} = 123.54$ kg

13. (JEE Main 2021 (Online) 26th August Evening Shift )

A cylindrical container of volume 4.0 $\times$ 10^{$-$ 3} m³ contains one mole of hydrogen and two moles of carbon dioxide. Assume the temperature of the mixture is 400 K. The pressure of the mixture of gases is :

[Take gas constant as 8.3 J mol^{$-$ 1} K^{$-$ 1}]

(A) 249 $\times$ 10¹ Pa

(B) 24.9 $\times$ 10³ Pa

(C) 24.9 $\times$ 10⁵ Pa

(D) 24.9 Pa

Correct answer is (C)

V = 4 $\times$ 10^{$-$ 3} m³

n = 3 moles

T = 400 K

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

P = $\frac{3 \times 8.3 \times 400}{4 \times 10^{- 3}}$

= 24.9 $\times$ 10⁵ Pa

14. (JEE Main 2021 (Online) 20th July Evening Shift )

Which of the following graphs represent the behavior of an ideal gas? Symbols have their usual meaning.

(A) JEE Main 2021 (Online) 20th July Evening Shift Physics - Heat and Thermodynamics Question 145 English Option 1

(B) JEE Main 2021 (Online) 20th July Evening Shift Physics - Heat and Thermodynamics Question 145 English Option 2

(C) JEE Main 2021 (Online) 20th July Evening Shift Physics - Heat and Thermodynamics Question 145 English Option 3

(D) JEE Main 2021 (Online) 20th July Evening Shift Physics - Heat and Thermodynamics Question 145 English Option 4

Correct answer is (C)

PV = nRT

PV $\propto$ T

Straight line with positive slope (nR)

15. (JEE Main 2021 (Online) 16th March Morning Shift )

The volume V of an enclosure contains a mixture of three gases, 16 g of oxygen, 28 g of nitrogen and 44 g of carbon dioxide at absolute temperature T. Consider R as universal gas constant. The pressure of the mixture of gases is :

(A) $\frac{3 R T}{V}$

(B) $\frac{4 R T}{V}$

(C) $\frac{88 R T}{V}$

(D) $\frac{5}{2} \frac{R T}{V}$

Correct answer is (D)

No. of moles of O₂ :
n₁ = $\frac{16}{32}$ = 0.5 mole

No. of moles of N₂ :
n₂ = $\frac{28}{28}$ = 1 mole

No. of moles of CO₂ :
n₃ = $\frac{44}{44}$ = 1 mole

Total no. of moles in container : n = n₁ + n₂ + n₃

$∴$ n = 0.5 + 1 + 1 = $\frac{5}{2}$ moles

Now; PV = nRT

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

$\Rightarrow$ P = $\frac{5}{2} \frac{R T}{V}$

Kinetic Theory of Gases