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Physics NEET MCQ
Quiz 1
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Q.1
One mole of an ideal gas at an initial temperature of T K does 6R joule of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is 5/3, the final temperature of gas will be
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(T + 2.4) K
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(T - 2.4) K
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(T + 4) K
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(T - 4) K
Explanation
Solution: for an adiabatic process, work done BY the gas is W = nR(Tᵢ − T_f)/(γ − 1). 6R = (1)(R)(T − T_f)/(5/3 − 1) = R(T − T_f)/(2/3) = (3/2)R(T − T_f) 6 = (3/2)(T − T_f) → T − T_f = 4 → T_f = T − 4.
Q.2
In which process, the PV-indicator diagram is a straight line parallel to volume axis?
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Isothermal
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Isobaric
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Irreversible
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Adiabatic
Explanation
A horizontal line on a P-V diagram means pressure stays constant while volume changes — that is exactly an isobaric (constant-pressure) process.
Q.3
If the temperature of the source is increased, the efficiency of a Carnot engine?
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Increases
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Decreases
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Remain constant
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First increases and then remain constant
Explanation
Carnot efficiency η = 1 − T_c/T_h. Increasing the hot reservoir (source) temperature T_h makes the ratio T_c/T_h smaller, so efficiency η increases.
Q.4
An ideal gas A and a real gas B have their volumes increased from V to 2V under isothermal conditions . The increase in internal energy
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Will be the same in both A and B
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Will be zero in both the gases
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Of B will be more than that of A
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Of B will be more than that of A
Explanation
For an IDEAL gas, internal energy depends only on temperature, so at constant temperature (isothermal), ΔU = 0 regardless of any volume change — that's true for gas A. For a REAL gas with attractive intermolecular forces (like B), internal energy also depends on volume/separation between molecules (captured by the −a/V term in the van der Waals internal-energy expression) — expanding at constant temperature increases the average molecular separation, which genuinely raises internal energy for a real gas with attraction. So the physically correct comparison is that ΔU of B is GREATER than ΔU of A (which stays at zero) — not that both are zero. (The option set in the source data repeats "Of B will be more than that of A" twice, suggesting some data corruption here; the marked "zero in both" answer doesn't match standard real-gas thermodynamics.)
Q.5
In a Carnot cycle, order of process is
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isothermal expansion, adiabatic expansion and adiabatic compression
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isothermal expansion, adiabatic compression and adiabatic expansion
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adiabatic expansion, isothermal expansion and adiabatic compression
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none of the above
Explanation
The standard Carnot cycle consists of four steps, in this order: (1) isothermal expansion, (2) adiabatic expansion, (3) isothermal compression, (4) adiabatic compression — alternating between isothermal and adiabatic stages, with two expansions and two compressions overall. (The listed options here appear garbled/incomplete versions of this sequence, likely from a scraping issue; the sequence above is the standard, correct one.)
Q.6
A diatomic ideal gas is compressed adiabatically to 1/32 of its initial volume. If the initial temperature of the gas is $T_i$ ( in kelvin) and the final temperature is $pT_i$. The value of p is
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1
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2
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3
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4
Explanation
For adiabatic compression, TV^(γ−1) stays constant. For a diatomic gas, γ = 7/5: TᵢVᵢ^(γ−1) = T_f(Vᵢ/32)^(γ−1) T_f = Tᵢ × 32^(γ−1) = Tᵢ × 32^0.4 Since 32 = 2⁵, 32^0.4 = 2^(5×0.4) = 2² = 4. So T_f = 4Tᵢ, i.e. p = 4.
Q.7
In a given process on an ideal gas, dW=0 and dQ
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$\Delta T > 0$
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$\Delta T < 0$
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$\Delta T =0$
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Insufficient information
Explanation
(The question text is missing a "0" — it should read dQ < 0.) By the first law, dU = dQ − dW = dQ − 0 = dQ. Since dQ < 0, dU < 0 too, and for an ideal gas internal energy change tracks temperature change directly — so ΔT < 0 (the gas cools).
Q.8
During adiabatic compression of a gas, its temperature
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Falls
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Remains constant
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Rises
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Becomes zero
Explanation
During adiabatic compression, no heat enters or leaves the gas (Q = 0), so all the work done ON the gas by compressing it goes directly into raising its internal energy — which means its temperature rises.
Q.9
A given mass of a gas is compressed isothermally until its pressure is doubled. It is then allowed to expand adiabatically until its original volume is restored and its pressure is then found to be 0.75 of its initial pressure. The ratio of the specific heats of the gas is approximately
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1.20
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1.41
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1.67
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1.83
Explanation
Isothermal compression to double the pressure: P → 2P, V → V/2 (Boyle's law). Then adiabatic expansion back to the original volume V, ending at pressure 0.75P. Using PV^γ = constant between these last two states: (2P)(V/2)^γ = (0.75P)(V)^γ 2 × (1/2)^γ = 0.75 2^(1−γ) = 0.75 (1−γ) = ln(0.75)/ln(2) ≈ −0.415 γ ≈ 1.415 ≈ 1.41.
Q.10
“Heat cannot by itself flow from a body at a lower temperature to a body at higher temperature” is a statement of the consequence of
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the second law of thermodynamics
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conservation of momentum
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conservation of mass
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the first law of thermodynamics.
Explanation
The statement that heat never spontaneously flows from a colder body to a hotter one (without external work being done) is the Clausius formulation of the second law of thermodynamics.
Q.11
In the following questions, a statement of assertion is followed by a statement of reason. You are required to choose the correct one out of the given four responses and mark it as (a) If both assertion and reason are true and reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not correct explanation of the assertion. (c) If assertion is true, but reason is false. (d) If both assertion and reason are false. (e) If reason is true but assertion is false Assertion: In an isolated system, the entropy increases Reason : The processes in an isolated system are adiabatic
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(a)
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(b)
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(c)
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(d)
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(e)
Explanation
The Assertion is true (entropy of an isolated system increases for any irreversible process inside it — the core statement of the second law). The Reason is also true on its own — an isolated system exchanges no heat with its surroundings, so any process within it is indeed adiabatic (Q = 0). But being merely adiabatic doesn't by itself guarantee entropy increases — a REVERSIBLE adiabatic process has ΔS = 0. It's specifically the IRREVERSIBILITY of real processes inside the isolated system that drives entropy up, not adiabaticity alone. So the Reason doesn't correctly explain the Assertion.
Q.12
In the following questions, a statement of assertion is followed by a statement of reason. You are required to choose the correct one out of the given four responses and mark it as (a) If both assertion and reason are true and reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not correct explanation of the assertion. (c) If assertion is true, but reason is false. (d) If both assertion and reason are false. (e) If reason is true but assertion is false Assertion: The Carnot cycle is useful in understanding the performance of heat engines. Reason: The Carnot cycle provides a way of determining the maximum possible efficiency achievable with reservoirs of given temperatures
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(a)
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(b)
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(c)
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(d)
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(e)
Explanation
Both statements are true, and directly connected: the Carnot cycle sets the theoretical upper limit on efficiency achievable by any heat engine operating between two given temperatures — which is exactly why studying it is useful for understanding and benchmarking real heat engines' performance.
Q.13
In the following questions, a statement of assertion is followed by a statement of reason. You are required to choose the correct one out of the given four responses and mark it as (a) If both assertion and reason are true and reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not correct explanation of the assertion. (c) If assertion is true, but reason is false. (d) If both assertion and reason are false. (e) If reason is true but assertion is false Assertion: Air quickly leaking out of a balloon becomes cooler. Reason: The leaking air undergoes adiabatic expansion.
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(a)
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(b)
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(c)
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(d)
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(e)
Explanation
When gas escapes a balloon quickly, there isn't enough time for heat to be exchanged with the surroundings, so the expansion is effectively adiabatic. In an adiabatic expansion, the gas does work using its own internal energy (with no heat coming in to replace it), so its temperature — and that of the escaping air — drops, making it feel cooler.
Q.14
In the following questions, a statement of assertion is followed by a statement of reason. You are required to choose the correct one out of the given four responses and mark it as| (a) If both assertion and reason are true and reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not correct explanation of the assertion. (c) If assertion is true, but reason is false. (d) If both assertion and reason are false. (e) If reason is true but assertion is false Assertion: The temperature of the surface of the sun is approx. 6000K. if we make a big lens and focus the sun rays, we can produce a temperature of 8000K Reason: The higher temperature can be produced accordingly to the second law of thermodynamics
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(a)
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(b)
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(c)
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(d)
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(e)
Explanation
Both statements are false. Focusing sunlight with a lens can concentrate its intensity, but by fundamental thermodynamic principles (related to conservation of étendue/radiance in passive optical systems), it can never make the focused spot hotter than the source itself — so reaching 8000 K from a 6000 K source is not physically possible with a simple lens. The Reason is also wrong: the second law doesn't provide a way to exceed the source temperature this way — if anything, it's exactly what forbids it.
Q.15
An ideal gas having initial pressure P, volume V and temperature T is allowed to expand adiabatically until its volume becomes 5.66V while its temperature falls to T/How many degrees of freedom do the gas molecules have?
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5
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6
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8
Explanation
Using TV^(γ−1) = constant with T_f = T/2 and V_f = 5.66V: T·V^(γ−1) = (T/2)·(5.66V)^(γ−1) 2 = 5.66^(γ−1) γ − 1 = ln2/ln(5.66) ≈ 0.4 → γ ≈ 1.4 For a gas with f degrees of freedom, γ = 1 + 2/f → f = 2/(γ−1) = 2/0.4 = 5 (consistent with a diatomic gas).
Q.16
Obtain the workdone by a gas dring the expansion as a function of the initial pressure P and Volume V
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3/2 PV
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5/4 PV
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1/2 PV
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3/4 PV
Explanation
Continuing the same adiabatic process as the previous question (initial state P, V, T; final state 5.66V, T/2; γ = 1.4, using nR = PV/T): W = nR(Tᵢ − T_f)/(γ−1) = (PV/T)×(T − T/2)/(0.4) = (PV/T)×(T/2)/0.4 = (PV/2)/0.4 = 1.25PV = 5/4 PV.
Q.17
A monotonic gas is taken through ABCDA cycle. A->B Constant Pressure process B->C Constant volume process C->D Constant Pressure process D->A Constant volume process The PV coordinates of point A are (P,V) The PV coordinates of point B are (P,3V) The PV coordinates of point C are (3P,3V) The PV coordinates of point D are (3P,V) then which of the following is not true
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total work done in the cycle is -4V
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Change in Internal energy =0
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total work done in the cycle is 4V
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Work done in Process CD=-6PV
Explanation
Work done (∫P dV) in each leg of the cycle: A→B (const. P, V: V→3V): W = P×(3V−V) = 2PV B→C (const. V): W = 0 C→D (const. P=3P, V: 3V→V): W = 3P×(V−3V) = −6PV D→A (const. V): W = 0 Total work over the cycle = 2PV + 0 − 6PV + 0 = −4PV, not +4PV — so "total work done in the cycle is 4V [i.e. +4PV]" is the statement that is NOT true (the correct total is −4PV). Change in internal energy over a full cycle is indeed always zero (it returns to the same state), and the work in leg C→D is indeed −6PV, exactly as those other options state.
Q.18
The first law of thermodynamics dU=dQ-dW,indicates that when a system goes from its initial state to a final state
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dU is same for every path
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dQ is same for every path
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dQ+dW is same for every path
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dQ-2dW is same for every path
Explanation
Internal energy U is a state function — it depends only on the system's current state, not on how it got there — so dU (equivalently ΔU) between two given states is exactly the same no matter which path connects them. dQ and dW individually are NOT state functions; they depend on the specific path taken, even between the same two endpoints.
Q.19
A gas is contained in a vertical, friction less piston cylinder device. The piston has a mass of 20 kg with a cross-sectional area of 20 cm2 and is pulled with a force of 100N. If the atmospheric pressure is 100 kPa, determine the pressure inside.
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110 kPa
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100kPa
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150kPa
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200 Kpa
Explanation
Using a force balance on the piston (gas pressure pushing up from below, balanced by atmospheric pressure and the piston's weight pushing down, with the applied 100 N force helping support it): P_gas × A + F = P_atm × A + mg P_gas = P_atm + (mg − F)/A With m = 20 kg, g ≈ 9.8 m/s², A = 20 cm² = 0.002 m², F = 100 N: P_gas = 100 + (196 − 100)/0.002 (in Pa, then converted) ≈ 100 + 48 = 148 kPa. (This standard force-balance method gives roughly 148 kPa with the numbers as given, rather than exactly matching the marked 100 kPa — there may be a units or value mismatch in the source data. The method above is the correct general approach for this kind of piston-force-balance problem.)
Q.20
For an adiabatic process
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$\Delta S=0$
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$\Delta U=0$
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Q =0
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W=0
Explanation
For any adiabatic process, Q = 0 is the literal definition — that always holds, whether the process is reversible or not. ΔS = 0 holds specifically when the adiabatic process is also reversible (a "reversible adiabatic" process is exactly what's meant by "isentropic") — an irreversible adiabatic process can still have ΔS > 0 even with zero heat transfer. Basic-level problems on adiabatic processes conventionally assume the reversible, quasi-static case by default, which is the convention behind marking ΔS = 0 as the answer here; strictly speaking, Q = 0 is the more universally guaranteed property of any adiabatic process. (ΔU = 0 and W = 0 are not generally true for adiabatic processes — an adiabatic process can do nonzero work, changing internal energy via that work alone.)
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