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Physics NEET Questions — Free MCQ Practice

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

A.

$\frac{2}{5}$

B.

$\frac{3}{5}$

C.

$\frac{3}{7}$

D.

$\frac{5}{7}$

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Explanation

Fraction of supplied energy which increases the internal energy is given by

$f = \frac{Δ U}{{(Δ Q)}_{P}} = \frac{{(Δ Q)}_{V}}{{(Δ Q)}_{P}} = \frac{μ C_{V} Δ T}{μ C_{P} Δ T} = \frac{1}{γ}$

For diatomic gas $γ = \frac{7}{5}$ ⇒ $f = \frac{5}{7}$

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A monoatomic ideal gas, initially at temperature T₁, is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T₂ by releasing the piston suddenly. If L₁ and L₂are the lengths of the gas column before and after expansion respectively, then T₁/ T₂ is given by -

A.

${(\frac{L_{1}}{L_{2}})}^{2 / 3}$

B.

$\frac{L_{1}}{L_{2}}$

C.

$\frac{L_{2}}{L_{1}}$

D.

${(\frac{L_{2}}{L_{1}})}^{2 / 3}$

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Explanation

$T_{1} {V_{1}}^{γ - 1} = T_{2} {V_{2}}^{γ - 1}$

$\Rightarrow \frac{T_{1}}{T_{2}} = {(\frac{V_{2}}{V_{1}})}^{γ - 1}$ $= {(\frac{L_{2} A}{L_{1} A})}^{\frac{5}{3} - 1} = {(\frac{L_{2}}{L_{1}})}^{\frac{2}{3}}$

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A mono atomic gas is supplied the heat Q very slowly keeping the pressure constant. The work done by the gas will be

A.

$\frac{2}{3} Q$

B.

$\frac{3}{5} Q$

C.

$\frac{2}{5} Q$

D.

$\frac{1}{5} Q$

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Explanation

$Δ Q = Δ U + Δ W$ ⇒ $Δ W = {(Δ Q)}_{P} - Δ U = {(Δ Q)}_{P} [1 - \frac{{(Δ Q)}_{V}}{{(Δ Q)}_{P}}]$

$= {(Δ Q)}_{P} [1 - \frac{C_{V}}{C_{P}}] = Q = [1 - \frac{3}{5}] = \frac{2}{5} Q$

âˆµ ${(Δ Q)}_{P} = Q$ and $γ = \frac{5}{3}$ for monatomic gas

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A gas mixture consists of 2 moles of oxygen and 4 moles argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system is

A.

4 RT

B.

15 RT

C.

9 RT

D.

11 RT

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Explanation

Oxygen is diatomic gas, hence its energy of two moles $= 2 \times \frac{5}{2} R T = 5 R T$

Argon is a monoatomic gas, hence its internal energy of 4 moles $= 4 \times \frac{3}{2} R T = 6 R T$

Total Internal energy = (6 + 5)RT = 11RT

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An ideal gas expands isothermally from a volume V₁ to V₂ and then compressed to original volume V₁adiabatically. Initial pressure is P₁ and final pressure is P₃. The total work done is W. Then -

A.

$P_{3} > P_{1}, W > 0$

B.

$P_{3} < P_{1}, W < 0$

C.

$P_{3} > P_{1}, W < 0$

D.

$P_{3} = P_{1}, W = 0$

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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

Physics MCQs for NEET — Practice Questions with Answers

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