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For an isothermal expansion of a perfect gas, the value of $\frac{Δ P}{P}$ is equal -

$- γ^{1 / 2} \frac{Δ V}{V}$

$- \frac{Δ V}{V}$

$- γ \frac{Δ V}{V}$

$- γ^{2} \frac{Δ V}{V}$

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Explanation

Differentiate PV = constant w.r.t V

⇒ $P Δ V + V Δ P = 0$ ⇒ $\frac{Δ P}{P} = - \frac{Δ V}{V}$

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If a gas is heated at constant pressure, its isothermal compressibility

Remains constant

Increases linearly with temperature

Decreases linearly with temperature

Decreases inversely with temperature

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Explanation

$E_{θ} = P$ , if P = constant, $E_{θ} =$ constant

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In isothermal expansion, the pressure is determined by -

Temperature only

Compressibility only

Both temperature and compressibility

None of these

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Explanation

For such a case, pressure $= \frac{1}{Compressibility}$

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The isothermal bulk modulus of a perfect gas at normal pressure is -

$1 .013 \times 10^{5} N / m^{2}$

$1 .013 \times 10^{6} N / m^{2}$

$1.013 \times 10^{- 11} N / m^{2}$

$1 .013 \times 10^{11} N / m^{2}$

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Explanation

$E_{θ} = P = 1 .013 \times 10^{5} N / m^{2}$

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In an isothermal change, an ideal gas obeys -

Boyle's law

Charle's law

Gaylussac law

None of the above

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Explanation

In isothermal process, compressibility E_θ = ρ.

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An ideal gas A and a real gas B have their volumes increased from V to 2 V under isothermal conditions. The increase in internal energy

Will be same in both A and B

Will be zero in both the gases

Of B will be more than that of A

Of A will be more than that of B

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Explanation

No change in the internal energy of ideal gas but for real gas, internal energy increases because work is done against intermolecular forces.

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The specific heat of a gas in an isothermal process is -

Infinite

Zero

Negative

Remains constant

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Explanation

In isothermal process temperature remains constant. i.e., $Δ T = 0$ .

Hence according to $C = \frac{Q}{m Δ T}$ ⇒ $C_{i s o} = \infty$

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The latent heat of vaporisation of water is 2240 J/gm. If the work done in the process of expansion of 1 g is 168 J, then increase in internal energy is

2408 J

2240 J

2072 J

1904 J

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Explanation

$Δ Q = Δ U + Δ W$

$\Rightarrow Δ U = Δ Q - Δ W = 2240 - 168 = 2072 J$

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A cylinder fitted with a piston contains 0.2 moles of air at temperature 27°C. The piston is pushed so slowly that the air within the cylinder remains in thermal equilibrium with the surroundings. Find the approximate work done by the system if the final volume is twice the initial volume

543 J

345 J

453 J

600 J

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Explanation

$W = μ R T \log_{e} (\frac{V_{2}}{V_{1}})$

$= 0 .2 \times 8 .3 \times \log_{e} 2 \times (27 + 273)$

$= 0 .2 \times 8 .3 \times 300 \times 0 .693 = 345 J$

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If a cylinder containing a gas at high pressure explodes, the gas undergoes -

Reversible adiabatic change and fall of temperature

Reversible adiabatic change and rise of temperature

Irreversible adiabatic change and fall of temperature

Irreversible adiabatic change and rise of temperature

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Explanation

Gas cylinder suddenly explodes is an irreversible adiabatic change and work done against expansion reduces the temperature.

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