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NEET NEET Questions — Free MCQ Practice

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Work done by a system under isothermal change from a volume V₁ to V₂ for a gas which obeys Vander Waal's equation $(V - β n) (P + \frac{α n^{2}}{V}) = n R T$

A.

$n R T \log_{e} (\frac{V_{2} - n β}{V_{1} - n β}) + α n^{2} (\frac{V_{1} - V_{2}}{V_{1} V_{2}})$

B.

$n R T \log_{10} (\frac{V_{2} - α β}{V_{1} - α β}) + α n^{2} (\frac{V_{1} - V_{2}}{V_{1} V_{2}})$

C.

$n R T \log_{e} (\frac{V_{2} - n α}{V_{1} - n α}) + β n^{2} (\frac{V_{1} - V_{2}}{V_{1} V_{2}})$

D.

$n R T \log_{e} (\frac{V_{1} - n β}{V_{2} - n β}) + α n^{2} (\frac{V_{1} V_{2}}{V_{1} - V_{2}})$

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Explanation

According to given Vander Waal’s equation

$P = \frac{n R T}{V - n β} - \frac{α n^{2}}{V^{2}}$

Work done, $W = \int_{V_{1}}^{V_{2}} P d V = n R T \int_{V_{1}}^{V_{2}} \frac{d V}{V - n β} - α n^{2} \int_{V_{1}}^{V_{2}} \frac{d V}{V^{2}}$

$= n R T [\log_{e} (V - n β)]_{V_{1}}^{V_{2}} + α n^{2} {[\frac{1}{V}]}_{V_{1}}^{V_{2}}$

$= n R T \log_{e} \frac{V_{2} - n β}{V_{1} - n β} + α n^{2} (\frac{V_{1} - V_{2}}{V_{1} V_{2}})$

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The molar heat capacity in a process of a diatomic gas if it does a work of $\frac{Q}{4}$ when a heat of Q is supplied to it is -

A.

$\frac{2}{5} R$

B.

$\frac{5}{2} R$

C.

$\frac{10}{3} R$

D.

$\frac{6}{7} R$

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Explanation

$d U = C_{V} d T = (\frac{5}{2} R) d T$ or $d T = \frac{2 (d U)}{5 R}$ …..(i)

From first law of thermodynamics

$d U = d Q - d W = Q - \frac{Q}{4} = \frac{3 Q}{4}$ .

Now molar heat capacity $C = \frac{d Q}{d T} = \frac{Q}{\frac{2 (d U)}{5 R}} = \frac{5 R Q}{2 (\frac{3 Q}{4})} = \frac{10}{3} R$ .

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An insulator container contains 4 moles of an ideal diatomic gas at temperature T. Heat Q is supplied to this gas, due to which 2 moles of the gas are dissociated into atoms but temperature of the gas remains constant. Then

A.

Q = 2RT

B.

Q = RT

C.

Q = 3RT

D.

Q = 4RT

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Explanation

Q = ΔU = U_f – U_i = [internal energy of 4 moles of a monoatomic gas + internal energy of 2 moles of a diatomic gas] – [internal energy of 4 moles of a diatomic gas]

$= (4 \times \frac{3}{2} R T + 2 \times \frac{5}{2} R T) - (4 \times \frac{5}{2} R T)$ = RT

= RT

Note : (1) 2 moles of diatomic gas becomes 4 moles of a monoatomic gas when gas dissociated into atoms.

Internal energy of μ moles of an ideal gas of degrees of freedom F is given by $U = \frac{f}{2} μ R T$

f = 3 for a monoatomic gas and 5 for diatomic gas.

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The volume of air increases by 5% in its adiabatic expansion. The percentage decrease in its pressure will be -

A.

5%

B.

6%

C.

7%

D.

8%

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Explanation

$P V^{γ} = K$ or $P γ V^{γ - 1} d V + d P . V^{γ} = 0$

or $\frac{d P}{P} = - γ \frac{d V}{V}$ or $\frac{d P}{P} \times 100 = - γ (\frac{d V}{V} \times 100)$

= –1.4 × 5 = 7%

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The temperature of a hypothetical gas increases to $\sqrt{2}$ times when compressed adiabatically to half the volume. Its equation can be written as

A.

PV^3/2 = constant

B.

PV^5/2 = constant

C.

PV^7/3 = constant

D.

PV^4/3 = constant

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Explanation

$T V^{γ - 1}$ = constant

∴ $\frac{T_{1}}{T_{2}} = {(\frac{V_{2}}{V_{1}})}^{γ - 1}$ or ${(\frac{1}{2})}^{γ - 1} = \sqrt{\frac{1}{2}}$

∴ $γ - 1 = \frac{1}{2}$ or $γ = \frac{3}{2}$

∴ PV^3/2= constant

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Two Carnot engines A and B are operated in succession. The first one, A receives heat from a source at T₁ = 800 K and rejects to sink at T₂K. The second engine B receives heat rejected by the first engine and rejects to another sink at T₃ = 300 K. If the work outputs of two engines are equal, then the value of T₂ is -

A.

100K

B.

300K

C.

550K

D.

700K

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Explanation

$η_{A} = \frac{T_{1} - T_{2}}{T_{1}} = \frac{W_{A}}{Q_{1}}$ ⇒ $η_{B} = \frac{T_{2} - T_{3}}{T_{2}} = \frac{W_{B}}{Q_{2}}$

∴ $\frac{Q_{1}}{Q_{2}} = \frac{T_{1}}{T_{2}} \times \frac{T_{2} - T_{3}}{T_{1} - T_{2}} = \frac{T_{1}}{T_{2}}$

∴ W_A = W_B

∴ $T_{2} = \frac{T_{1} + T_{3}}{2} = \frac{800 + 300}{2} = 550 K$

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When an ideal monoatomic gas is heated at constant pressure, fraction of heat energy supplied which increases the internal energy of gas, is

A.

$\frac{2}{5}$

B.

$\frac{3}{5}$

C.

$\frac{3}{7}$

D.

$\frac{3}{4}$

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Explanation

For monoatomic gas

$γ = \frac{C_{P}}{C_{V}} = \frac{5}{3}$ we know $Δ Q = n C_{P} Δ T$

and $Δ U = n C_{V} Δ T$ ⇒ $\frac{Δ U}{Δ Q} = \frac{C_{V}}{C_{P}} = \frac{3}{5}$

i.e. fraction of heat energy to increase the internal energy be 3/5.

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When an ideal gas (γ = 5/3) is heated under constant pressure, then what percentage of given heat energy will be utilised in doing external work ?

A.

40 %

B.

30 %

C.

60 %

D.

20 %

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Explanation

$Δ Q = Δ U + Δ W \Rightarrow \frac{Δ W}{Δ Q} = 1 - \frac{Δ U}{Δ Q} = 1 - \frac{n C_{V} d T}{n C_{P} d T}$

$\Rightarrow \frac{Δ W}{Δ Q} = 1 - \frac{C_{V}}{C_{P}} = 1 - \frac{3}{5} = \frac{2}{5} = 0 .4$

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Which one of the following gases possesses the largest internal energy?

A.

2 moles of helium occupying 1m³at 300 K

B.

56 kg of nitrogen at $10^{5}$ Nm^–2 and 300 K

C.

8 grams of oxygen at 8 atm and 300 K

D.

6 × 10²⁶ molecules of argon occupying 40 m³ at 900 K

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Explanation

$U = \frac{n fR T}{2}$

⇒ ${(U)}_{N} = \frac{56 \times 10^{3}}{28} \times \frac{5}{2} R \times 300$

and ${(U)}_{A r} = \frac{6 \times 10^{26}}{6 \times 10^{23}} \times \frac{3}{2} R \times 900 \Rightarrow {(U)}_{A r} < {(U)}_{N}$

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Two samples A and B of a gas initially at the same pressure and temperature are compressed from volume V to V/2 (A isothermally and B adiabatically). The final pressure of A is

A.

Greater than the final pressure of B

B.

Equal to the final pressure of B

C.

Less than the final pressure of B

D.

Twice the final pressure of B

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Explanation

A is compressed isothermally, hence

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

and B is compressed adiabatically, hence

$P_{1} V^{γ} = P_{2} {(\frac{V}{2})}^{γ} \Rightarrow P_{2} = {(2)}^{γ} P_{1}$

Since $γ > 1$ , hence $P_{2}' > P_{2}$ or $P_{2} < {P^{'}}_{2}$

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