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Initial pressure and volume of a gas are P and V respectively. First it is expanded isothermally to volume 4V and then compressed adiabatically to volume V. The final pressure of gas will be [Given : $γ = 1.5]$ -

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Explanation

In isothermal process $P_{1} V_{1} = P_{2} V_{2}$

or $P V = P_{2} \times 4 V$

∴ $P_{2} = \frac{P}{4}$

In adiabatic process

$P_{2} {V_{2}}^{γ} = P_{3} {V_{3}}^{γ}$

⇒ $\frac{P}{4} \times {(4 V)}^{1 .5} = P_{2} V^{1 .5}$

⇒ $P_{3} = 2 P$

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A reversible engine converts one-sixth of the heat input into work. When the temperature of the sink is reduced by 62°C, the efficiency of the engine is doubled. The temperatures of the source and sink are -

80°C, 37°C

95°C, 28°C

90°C, 37°C

99°C, 37°C

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Explanation

Initially $η = (1 - \frac{T_{2}}{T_{1}}) = \frac{W}{Q} = \frac{1}{6}$ ...(i)

Finally $η' = (1 - \frac{T_{2}'}{T_{1}}) = (1 - \frac{(T_{2} - 62)}{T_{1}}) = 1 - \frac{T_{2}}{T_{1}} + \frac{62}{T_{1}}$

$= η + \frac{62}{T_{1}}$ ....(ii)

It is given that $η' = 2 η .$ Hence solving equation (i) and (ii)

⇒ $T_{1} = 372 K = 99 ° C$ and $T_{2} = 310 K = 37 ° C$

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An ideal gas expands in such a manner that its pressure and volume can be related by equation $P V^{5 / 3} = c o n s \tan t$ . During this process, the gas is

Heated

Cooled

Neither heated nor cooled

First heated and then cooled

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Explanation

$P V^{5 / 3} =$ constant represents adiabatic equation.

So during the expansion of ideal gas internal energy of gas decreases and temperature falls.

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P-V diagram of a diatomic gas is a straight line passing through origin. The molar heat capacity of the gas in the process will be -

4 R

2.5 R

3 R

$\frac{4 R}{3}$

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Explanation

P-V diagram of the gas is a straight line passing through origin. Hence $P \propto V$ or $P V^{- 1} =$ constant

Molar heat capacity in the process $P V^{x} = constant is$

$C = \frac{R}{γ - 1} + \frac{R}{1 - x}$ ; Here $γ = 1 .4$ (For diatomic gas)

⇒ $C = \frac{R}{1 .4 - 1} + \frac{R}{1 + 1}$ ⇒ $C = 3 R$

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Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300 K. The piston of A is free to move while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30 K, then the rise in temperature of the gas in B is

30 K

18 K

50 K

42 K

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Explanation

In both cylinders A and B the gases are diatomic (γ = 1.4). Piston A is free to move i.e. it is isobaric process. Piston B is fixed i.e. it is isochoric process. If same amount of heat ΔQ is given to both then

${(Δ Q)}_{isobaric} = {(Δ Q)}_{isochoric}$ ⇒ $μ C_{p} {(Δ T)}_{A} = μ C_{v} {(Δ T)}_{B}$

⇒ ${(Δ T)}_{B} = \frac{C_{p}}{C_{v}} {(Δ T)}_{A} = γ {(Δ T)}_{A} = 1 .4 \times 30 = 42 K .$

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An isolated system-

is a specified region where transfers of energy and mass take place

is a region of constant mass and only energy is allowed to close the boundaries

is one in which mass within the system is not necessarily constant.

cannot transfer either energy of mass to or from the surroundings.

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Explanation

(D) An isolated system is one which is separated so that the mass and energy of the system cannot be transferred to or from the surroundings.

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The pressure and temperature of two different gases are P and T having the volume V for each. They are mixed keeping the same volume and temperature, the pressure of the mixture will be- [Pb. PMT 1997, 98; DPMT 1999; MH CET 2003]

P/2

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Explanation

$μ_{1} = \frac{P V}{R T}, μ_{2} = \frac{P V}{R T} P' = \frac{(μ_{1} + μ_{2}) R T}{V} = \frac{2 P V}{R T} \times \frac{R T}{V} = 2 P$

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If the mean free path of atoms is doubled then the pressure of the gas will become: [RPMT 2000]

P/4

P/2

P/8

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Explanation

Mean free path, $λ \propto \frac{1}{P};$ If $λ$ is doubled then P becomes half.

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A diatomic molecule has how many degrees of freedom- [Pb. PET 2000]

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Explanation

A diatomic molecule has three translational and two rotational degrees of freedom.

Hence, total degrees of freedom f = 3 + 2 = 5

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A thermally insulated piston divides a container into two compartments. Volume, temperature, and pressure in the right compartment are 2V, T, and 2P, while in the left compartment the respective values are V, T, and P. If the piston can slide freely, then in the final equilibrium position, the volume of the right-hand compartment will be-

$\frac{3 V}{5}$

$\frac{9 V}{4}$

$\frac{12 V}{5}$

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