Are these Thermodynamics questions free to attempt?

Yes. You can attempt every Thermodynamics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

Do I need an account to practise Thermodynamics MCQs?

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

Are these previous year NEET questions?

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.

Thermodynamics NEET Questions — Free MCQ Practice

● ● ● ● ●

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

You've reached today's free limit of 20 questions. Log in to keep practising for free.

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

● ● ● ● ●

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.

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

You've reached today's free limit of 20 questions. Log in to keep practising for free.

● ● ● ● ●

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.

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

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

According to given Vander Waal’s equation

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

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

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

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

● ● ● ● ●

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.

$\frac{6}{7} R$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

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

● ● ● ● ●

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

You've reached today's free limit of 20 questions. Log in to keep practising for free.

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.

● ● ● ● ●

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%

You've reached today's free limit of 20 questions. Log in to keep practising for free.

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%

● ● ● ● ●

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

You've reached today's free limit of 20 questions. Log in to keep practising for free.

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

● ● ● ● ●

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

You've reached today's free limit of 20 questions. Log in to keep practising for free.

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

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

● ● ● ● ●

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.

$\frac{3}{4}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

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.

● ● ● ● ●

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 %

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

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

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

Thermodynamics MCQs for NEET — Physics Questions with Answers

Frequently Asked Questions