Q.Figure 12.8 shows plot of $PV/T$ versus $P$ for $1.00\times10^{-3}$ kg of oxygen gas at two different temperatures.

Showing the 12 most recent of 19 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.An ideal gas of density 'ρ' is enclosed in a vessel of volume V at $0^\circ C$ temperature and $p_0$ atmospheric pressure. If the pressure inside the vessel decreased by Δp due to a leakage. The mass of the leaked out gas is (A) $V.\rho.p_0$ (B) $V\rho/p_0$ (C) $V\Delta\rho/p$ (D) $\dfrac{V.\rho.\Delta p}{p_0}$

   ›Reveal solutionSolution

   With volume and temperature both held fixed, the ideal gas law makes pressure directly proportional to the mass of gas inside — so a pressure drop $\Delta p$ corresponds to a proportional mass loss, and the known initial density supplies the proportionality constant. Answer: (D).

   Concept and Intuition

   The ideal gas equation $pV=\dfrac{m}{M_{molar}}RT$ can be rearranged as $p=\dfrac{m}{V}\cdot\dfrac{RT}{M_{molar}}$ — i.e., at fixed $V$ and $T$, pressure is directly proportional to the mass of gas enclosed (through density $=m/V$). This means we don't need to know the molar mass or $R$ explicitly at all: the given initial density $\rho$ at pressure $p_0$ already encodes the full proportionality constant between mass and pressure for this gas, in this vessel, at this temperature. A pressure drop $\Delta p$ then converts directly to a mass loss using that same ratio.

   Step-by-Step Solution

   1. Initial state: density $\rho=\dfrac{m_0}{V}$ at pressure $p_0$ (mass $m_0$, volume $V$, temperature $0^\circ C$ fixed).
   2. From the ideal gas law, since $V$ and $T$ don't change, $p\propto m$ — specifically $\dfrac{p}{m}=\dfrac{RT}{VM_{molar}}=$ constant. Using the initial values, this constant equals $\dfrac{p_0}{m_0}=\dfrac{p_0}{\rho V}$.
   3. After the leak, pressure drops by $\Delta p$, so mass drops by $\Delta m$, and the same proportionality holds for the change:

   $\frac{\Delta p}{\Delta m}=\frac{p_0}{\rho V} \quad\Rightarrow\quad \Delta m=\frac{\rho V\,\Delta p}{p_0}.$

   Common Mistakes

   • Trying to bring in the universal gas constant $R$ or molar mass explicitly — they're not needed since the initial density $\rho$ already packages that information for this specific gas/vessel/temperature.
   • Sign/ratio confusion — mass lost is proportional to pressure dropped, in the same ratio as the original mass was to the original pressure (not the final pressure).

   ✓Final answer

   The correct option is (D) — $\dfrac{V.\rho.\Delta p}{p_0}$.

   ANSWER: D

2. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.An ideal gas is expanding such that $P^2T = \text{constant}$. The coefficient of volume expansion of the gas is (P- Pressure, T- temperature of the gas) (A) $\frac{1}{T}$ (B) $\frac{2}{T}$ (C) $\frac{3}{T}$ (D) $\frac{3}{2T}$

   ›Reveal solutionSolution

   This tests combining the ideal gas law with a given process equation to find how $V$ depends on $T$ along that specific process, then differentiating to get the volume-expansion coefficient — answer $\frac{3}{2T}$.

   Concept and Intuition

   The "coefficient of volume expansion" $\gamma = \frac{1}{V}\left(\frac{dV}{dT}\right)$ is normally quoted at constant pressure, but here the gas follows a specific process ($P^2T =$ constant) where $P$ itself changes with $T$. So we must first eliminate $P$ using the ideal gas law to get $V$ purely as a function of $T$ along this process, and then differentiate that relation.

   Step-by-Step Solution

   1. Ideal gas law: $PV = nRT \implies P = \dfrac{nRT}{V}$.
   2. Given process: $P^2 T = C$ (constant). Substitute $P$:

   $\left(\frac{nRT}{V}\right)^2 T = C \implies \frac{n^2R^2 T^3}{V^2} = C$

   3. Solve for $V$ in terms of $T$:

   $V^2 = \frac{n^2R^2}{C}T^3 \implies V = \left(\frac{nR}{\sqrt{C}}\right)T^{3/2}$

   So $V \propto T^{3/2}$.

   4. Differentiate:

   $\frac{dV}{dT} = \frac{3}{2}\left(\frac{nR}{\sqrt C}\right)T^{1/2} = \frac{3}{2}\cdot\frac{V}{T}$

   5. Coefficient of volume expansion:

   $\gamma = \frac{1}{V}\frac{dV}{dT} = \frac{3}{2T}$

   Common Mistakes

   • Assuming $P$ is constant and using the standard $\gamma=1/T$ result for an isobaric ideal gas — that formula does not apply here because $P$ varies with $T$ along this process.
   • Sign/exponent slips when substituting $P=nRT/V$ into $P^2T=C$ (easy to drop the square).

   ✓Final answer

   The correct option is (D) — $\frac{3}{2T}$.

   ANSWER: D

3. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.The density ($\rho$) versus pressure (P) of a given mass of an ideal gas is shown at two temperatures $T_1$ and $T_2$. The relation between $T_1$ and $T_2$ is [FIGURE: graph of density $\rho$ (y-axis) vs pressure P (x-axis) showing two straight lines; the line labelled $T_1$ has a steeper slope (higher $\rho$ for a given P), and the line labelled $T_2$ has a smaller slope (lower $\rho$ for a given P)] (A) $T_1 > T_2$ (B) $T_2 > T_1$ (C) $T_1 = T_2$ (D) $T_1 > T_2$ or $T_2 > T_1$

   ›Reveal solutionSolution

   Tests how the slope of a $\rho$–$P$ graph for an ideal gas relates inversely to temperature. Since the steeper line ($T_1$) has the larger slope, it corresponds to the lower temperature, giving $T_2 > T_1$.

   Concept and Intuition

   Starting from the ideal gas law $PV=nRT$, and writing $n = m/M$ (mass $m$, molar mass $M$):

   $PV = \frac{m}{M}RT \implies P = \frac{m}{V}\cdot\frac{RT}{M} = \rho\frac{RT}{M} \implies \rho = \frac{M}{RT}\,P$

   So for a fixed gas (fixed $M$), a $\rho$ vs $P$ plot is a straight line through the origin with slope $\dfrac{M}{RT}$. This slope shrinks as temperature rises — physically, at higher temperature the same pressure supports a lower density (the gas is "puffier"/more expanded), which is exactly the ideal gas law's statement that hot gas at fixed pressure has lower density.

   Step-by-Step Solution

   1. From $PV=nRT$, derive $\rho = \dfrac{M}{RT}P$ — a straight line through the origin with slope $\propto \dfrac{1}{T}$.
   2. The graph shows $T_1$'s line steeper than $T_2$'s line (higher $\rho$ for the same $P$), i.e. slope$(T_1) >$ slope$(T_2)$.
   3. Since slope $\propto 1/T$, a larger slope means a smaller $T$: slope$(T_1)>$slope$(T_2) \implies T_1 < T_2$.
   4. Therefore $T_2 > T_1$.

   Common Mistakes

   • Assuming (incorrectly) that a steeper line means a higher temperature — it's the opposite here, because slope is inversely proportional to $T$.
   • Confusing this $\rho$-$P$ graph with a $P$-$V$ isotherm graph (where higher $T$ isotherms lie further from the origin) — the relationship is different because density and volume are inversely related.

   ✓Final answer

   The correct option is (B) — $T_2 > T_1$.

   ANSWER: B

4. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.A mass of an ideal gas of volume V at pressure P undergoes the cycle of changes shown in the graph. At which point is the gas coolest and hottest? [FIGURE] (a P–V graph, axes labelled $P/10^5\,\mathrm{Nm^{-2}}$ vertical and $V/10^{-4}\,\mathrm{m^3}$ horizontal; a right-triangle cycle with vertex X at (V=1, P=4), vertex Y at (V=5, P=1), and vertex Z at (V=1, P=1); the cycle runs X down to Y along the hypotenuse, Y back to Z along the bottom, and Z up to X along the left side) (A) coolest – X, hottest – Y (B) coolest – Y, hottest – X (C) coolest – Y, hottest – Z (D) coolest – Z, hottest – Y

   ›Reveal solutionSolution

   Using $T\propto PV$ from the ideal gas law at each vertex of the cycle, Z is coolest and Y is hottest.

   Concept and Intuition

   For a fixed amount of ideal gas, $PV=nRT$, so temperature is directly proportional to the product $PV$ at any state point on a P–V diagram. Comparing $PV$ at the three vertices of the cyclic process directly ranks their temperatures — no need to trace the path.

   Step-by-Step Solution

   1. At X: $P=4\times10^5\,\mathrm{Nm^{-2}}$ (in the graph's units, $P=4$), $V=1\times10^{-4}\,\mathrm{m^3}$ (units, $V=1$). So $PV \propto 4\times1=4$.
   2. At Y: $P=1$, $V=5$ (in the same graph units). $PV\propto1\times5=5$.
   3. At Z: $P=1$, $V=1$. $PV\propto1\times1=1$.
   4. Since $T\propto PV$: $T_Z(\propto1) < T_X(\propto4) < T_Y(\propto5)$.
   5. Therefore the coolest point is Z and the hottest point is Y.

   Common Mistakes

   • Assuming the point with highest pressure (X) must be hottest — temperature depends on the product $PV$, not pressure alone.
   • Reading the graph axes' scale factors incorrectly (both axes carry a $\times10^{\pm}$ multiplier that cancels out when just comparing ratios).

   ✓Final answer

   The correct option is (D) — coolest – Z, hottest – Y.

   ANSWER: D

5. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.The work done to increase the volume of 2 moles of an ideal gas from V to 2V at a constant temperature T is W. The work to be done to increase the volume of 2 moles of the same gas from 2V to 4V at the same constant temperature T is (A) 0.5W (B) W (C) 2W (D) 4W

   ›Reveal solutionSolution

   Isothermal work depends only on the ratio of final to initial volume, not their absolute values; since both steps double the volume, they require equal work. Answer: W.

   Concept and Intuition

   For an isothermal process, $W=nRT\ln(V_f/V_i)$. The key insight is that this depends only on the ratio $V_f/V_i$, not on the absolute volumes. Doubling the gas's volume always requires the same work at a given temperature, regardless of where you start.

   Step-by-Step Solution

   1. First process ($V\to2V$, $n=2$ mol): $W = 2RT\ln\left(\dfrac{2V}{V}\right)=2RT\ln2$.
   2. Second process ($2V\to4V$): $W' = 2RT\ln\left(\dfrac{4V}{2V}\right)=2RT\ln2$.
   3. Since $\ln2=\ln2$ in both cases, $W'=W$.

   Common Mistakes

   • Assuming the second step needs more work because the volumes involved are larger in absolute terms — but only the ratio matters for isothermal work.

   ✓Final answer

   The correct option is (B) — W.

   ANSWER: B

6. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.In a container of volume 16.62 $\mathrm{m^3}$ at $0°\mathrm{C}$ temperature, 2 moles of oxygen, 5 moles of nitrogen and 3 moles of hydrogen are present, then the pressure in the container is (Universal gas constant $= 8.31 \mathrm{~J~mol^{-1}~K^{-1}}$) (A) 1570 Pa (B) 1270 Pa (C) 1365 Pa (D) 2270 Pa

   ›Reveal solutionSolution

   This tests applying the ideal gas law to a mixture of gases, where the total pressure depends only on the total number of moles (Dalton's law is implicit); the answer is (C) 1365 Pa.

   Concept and Intuition

   For an ideal gas mixture in a fixed volume and temperature, the total pressure obeys $PV = n_{\text{total}}RT$, where $n_{\text{total}}$ is simply the sum of moles of all gases present — the identity of each gas doesn't matter for the total pressure, only the total mole count (this is Dalton's law in disguise).

   Step-by-Step Solution

   1. Total moles: $n = n_{O_2}+n_{N_2}+n_{H_2} = 2+5+3 = 10$ mol.
   2. Temperature: $T = 0°C = 273$ K.
   3. Ideal gas law: $P = \dfrac{nRT}{V} = \dfrac{10 \times 8.31 \times 273}{16.62}$.
   4. Numerator: $10\times 8.31 = 83.1$; $83.1 \times 273 = 22686.3$.
   5. $P = \dfrac{22686.3}{16.62} = 1365$ Pa.

   Common Mistakes

   • Computing partial pressures separately and forgetting to just sum the moles (equivalent, but more error-prone).
   • Using $T=0$ K instead of converting to Kelvin (273 K).

   ✓Final answer

   The correct option is (C) — 1365 Pa.

   ANSWER: C

7. AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
   Q.A vessel contains 8 g of air at 400 K. Some amount of air leaks out through the hole provided to it. After some time, pressure is halved and temperature is changed to 300 K. Find the mass of the air escaped (A) 5.33 g (B) 2.67 g (C) 6 g (D) 3.27 g

   ›Reveal solutionSolution

   With volume fixed, the mass of gas remaining is proportional to $P/T$; comparing the initial and final $P/T$ ratios gives the leaked mass as 2.67 g.

   Concept and Intuition

   For a rigid, sealed (except for the leak) container, $PV=nRT=\dfrac{m}{M}RT$. Since $V$, $R$, and molar mass $M$ don't change, the mass of gas present at any time is directly proportional to $P/T$ at that time. This lets us find the ratio of final to initial mass without knowing $M$ or $V$ explicitly.

   Step-by-Step Solution

   1. Initial state: mass $m_1=8$ g, at pressure $P$, temperature $T_1=400$ K.
   2. Final state: mass $m_2$ (unknown), pressure $P/2$, temperature $T_2=300$ K.
   3. Since $m\propto P/T$: $\dfrac{m_2}{m_1} = \dfrac{(P/2)/300}{P/400} = \dfrac{P/600}{P/400} = \dfrac{400}{600}=\dfrac23$.
   4. $m_2 = 8\times\dfrac23 = 5.33$ g.
   5. Mass escaped $= m_1-m_2 = 8-5.33 = 2.67$ g.

   Common Mistakes

   • Forgetting that pressure was HALVED (not just given as some new value $P_2$) when substituting.
   • Trying to use the ideal gas equation with absolute values of $P$ and $V$ (which aren't given) instead of working with the ratio, where $P$, $V$, $M$, $R$ all cancel out.

   ✓Final answer

   The correct option is (B) — 2.67 g.

   ANSWER: B

8. AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
   Q.A piston divides a closed gas cylinder into two parts. Initially the piston is kept pressed such that one part has pressure P and volume 5V and the other part has pressure 8P and volume V. The piston is now left free. The new pressure if the process is isothermal (A) $P_o = \dfrac{13}{6}P$ (B) $P_o = \dfrac{8}{13}P$ (C) $P_o = \dfrac{13}{8}P$ (D) $P_o = \dfrac{8}{6}P$

   ›Reveal solutionSolution

   Applying Boyle's Law separately to the gas on each side of the piston (each undergoes an isothermal change to reach the same final pressure), and using conservation of total volume, gives the equilibrium pressure $P_o = \tfrac{13}{6}P$.

   Concept and Intuition

   The piston is free to slide but the total cylinder volume is fixed. Since the process is isothermal, each trapped gas independently obeys $PV=\text{constant}$ as its volume changes, right up until the piston stops moving — at which point both sides must have the SAME pressure $P_o$ (otherwise there'd be a net force on the piston and it would keep moving).

   Step-by-Step Solution

   1. Side 1 initially: pressure $P$, volume $5V$. Applying Boyle's Law to its final state at pressure $P_o$: $P\times5V = P_o\times V_1' \Rightarrow V_1'=\dfrac{5PV}{P_o}$.
   2. Side 2 initially: pressure $8P$, volume $V$. Similarly: $8P\times V = P_o\times V_2' \Rightarrow V_2'=\dfrac{8PV}{P_o}$.
   3. Total volume is conserved (rigid cylinder): $V_1'+V_2' = 5V+V = 6V$.
   4. Substitute: $\dfrac{5PV}{P_o}+\dfrac{8PV}{P_o} = 6V \Rightarrow \dfrac{13PV}{P_o}=6V \Rightarrow P_o=\dfrac{13P}{6}$.

   Common Mistakes

   • Assuming the final pressure is just the average of the initial pressures (which is only true if the volumes were also equal) — it is not, because the two sides have different volumes.
   • Forgetting that the total volume (not each side's individual volume) is what's conserved.

   ✓Final answer

   The correct option is (A) — $P_o=\dfrac{13}{6}P$.

   ANSWER: A

9. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.Two ideal gases A and B of same number of moles expand at constant temperatures $T_1$ and $T_2$ respectively such that the pressure of gas A decreases by 50% and the pressure of gas B decreases by 75%. If the work done by both the gases is same, then $T_1 : T_2$ (A) 1 : 3 (B) 2 : 3 (C) 3 : 4 (D) 2 : 1

   ›Reveal solutionSolution

   Isothermal expansion work depends logarithmically on the pressure (or volume) ratio; equating the two given works pins down the temperature ratio.

   Concept and Intuition

   For an isothermal process on an ideal gas, $PV=nRT=\text{constant}$, so as pressure falls the volume must rise proportionally in the reciprocal sense: $V_f/V_i = P_i/P_f$. The work done in an isothermal expansion is $W=nRT\ln(V_f/V_i)$, which can therefore be rewritten directly in terms of the pressure ratio — useful because the problem gives percentage pressure drops, not volume ratios.

   Step-by-Step Solution

   1. Gas A: pressure decreases by 50%, so $P_{f,A} = 0.5\,P_{i,A}$, meaning $P_i/P_f = 2$.
   2. Work by A (n moles, temperature $T_1$): $W_A = nRT_1 \ln(P_i/P_f) = nRT_1\ln 2$.
   3. Gas B: pressure decreases by 75%, so $P_{f,B} = 0.25\,P_{i,B}$, meaning $P_i/P_f = 4$.
   4. Work by B (n moles, temperature $T_2$): $W_B = nRT_2\ln 4 = nRT_2 (2\ln 2) = 2nRT_2\ln 2$.
   5. Given $W_A = W_B$: $nRT_1\ln 2 = 2nRT_2\ln 2 \Rightarrow T_1 = 2T_2$.
   6. Hence $T_1:T_2 = 2:1$.

   Common Mistakes

   • Treating the pressure drop as directly giving the volume ratio without inverting ($V_f/V_i = P_i/P_f$, not $P_f/P_i$).
   • Forgetting $\ln 4 = 2\ln 2$ and mishandling the log algebra.

   ✓Final answer

   The correct option is (D) — 2 : 1.

   ANSWER: D

10. AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
    Q.If the percentage increase in the pressure of a given mass of gas in a container at constant temperature is 100%, then its volume (A) decreases by 100% (B) decreases by 50% (C) does not change (D) increases by 100%

    ›Reveal solutionSolution

    Boyle's Law at constant temperature: doubling the pressure halves the volume, a 50% decrease.

    Concept and Intuition

    At constant temperature, for a fixed mass of an ideal gas, pressure and volume are inversely proportional ($PV = \text{constant}$). Any percentage increase in one directly determines the reciprocal change in the other — it is not a symmetric percentage relationship.

    Step-by-Step Solution

    1. Boyle's Law: $P_1 V_1 = P_2 V_2$.
    2. Pressure increases by 100%, so $P_2 = 2P_1$.
    3. Then $V_2 = \dfrac{P_1 V_1}{P_2} = \dfrac{P_1 V_1}{2P_1} = \dfrac{V_1}{2}$.
    4. $V_2$ is half of $V_1$, i.e. the volume has decreased by 50% (not 100%, since volume can never become negative or zero here).

    Common Mistakes

    • Assuming a 100% increase in pressure causes a 100% decrease in volume (that would make volume zero, which is wrong — inverse proportionality is multiplicative, not additive).

    ✓Final answer

    The correct option is (B) — decreases by 50%.

    ANSWER: B

11. AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
    Q.In the given volume (V) - absolute temperature (T) graph of an ideal gas, the relation between the pressures 'P$_1$' and 'P$_2$' is [FIGURE] (a V-T graph with two straight lines through the origin; the P$_2$ line makes a 60° angle with the T-axis and the P$_1$ line makes a 30° angle with the T-axis) (A) $P_1 = \dfrac{P_2}{2}$ (B) $P_1 = \dfrac{P_2}{3}$ (C) $P_1 = 3P_2$ (D) $P_1 = 2P_2$

    ›Reveal solutionSolution

    The slope of a V–T graph for an ideal gas is proportional to $nR/P$. Since the slope is $\tan\theta$ (with $\theta$ measured from the T‑axis), the ratio of pressures is the inverse ratio of the tangents. Here $P_1 = 3P_2$, so option (C) is correct.

    ---

    The key is the Ideal Gas Law: $PV = nRT$. For a fixed amount of gas (constant $n$), if we plot volume $V$ against absolute temperature $T$, the equation becomes $V = \left(\frac{nR}{P}\right)T$. This is a straight line through the origin with slope $m = \frac{nR}{P}$.

    So the slope of the V–T line is inversely proportional to the pressure: a steeper line means a smaller pressure, and a shallower line means a larger pressure. The problem gives the angles the lines make with the T‑axis, so we can find the slopes from those angles.

    ---

    1. Interpret the angles correctly

       The T‑axis is horizontal. The angle a line makes with the horizontal is the same as the angle whose tangent gives the slope.

       • For $P_2$: angle = $60^\circ$ → slope $m_2 = \tan 60^\circ = \sqrt{3}$.
       • For $P_1$: angle = $30^\circ$ → slope $m_1 = \tan 30^\circ = \frac{1}{\sqrt{3}}$.

    2. Relate slope to pressure

       From $V = \frac{nR}{P}\,T$, we have $m = \frac{nR}{P}$. Therefore:

    $m_1 = \frac{nR}{P_1}, \quad m_2 = \frac{nR}{P_2}.$

    3. Take the ratio

    $\frac{m_1}{m_2} = \frac{nR/P_1}{nR/P_2} = \frac{P_2}{P_1}.$

    Substitute the slopes:

    $\frac{1/\sqrt{3}}{\sqrt{3}} = \frac{P_2}{P_1} \quad\Rightarrow\quad \frac{1}{3} = \frac{P_2}{P_1}.$

    Hence $P_1 = 3P_2$.

    Watch out

    A common mistake is to think the larger angle means larger pressure. But slope is $\tan\theta$, and $\tan 60^\circ > \tan 30^\circ$, so the steeper line ($P_2$) actually has lower pressure. Always check: slope $\propto 1/P$.

    Tip

    If you remember that for an ideal gas at constant pressure, $V \propto T$, then the line’s slope is directly $nR/P$. So comparing slopes is a quick way to compare pressures without re‑deriving.

    ✓Final answer

    The correct option is (C).

    ANSWER: C

12. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
    Q.In the given pressure (P) – absolute temperature (T) graph of an ideal gas, the relation between volumes $V_1, V_2, V_3$, and $V_4$ is [FIGURE] (a P-T graph with four straight lines through the origin labelled $V_1, V_2, V_3, V_4$ from steepest to shallowest slope, with dashed vertical reference lines at temperatures 100, 200, 300, 400 K) (A) $V_1=V_2=V_3=V_4$ (B) $V_1>V_2>V_3>V_4$ (C) $V_1>V_2>V_3<V_4$ (D) $V_1<V_2<V_3<V_4$

    ›Reveal solutionSolution

    For an ideal gas, the slope of a P–T line through the origin is proportional to $nR/V$; a steeper slope means a smaller volume. Since the slopes satisfy $V_1 > V_2 > V_3 > V_4$ (steepest to shallowest), the volumes obey $V_1 < V_2 < V_3 < V_4$.

    The key to this problem is the Ideal Gas Law, $PV = nRT$. When we plot pressure $P$ against absolute temperature $T$ for a fixed amount of gas, the relationship is linear if the volume $V$ is constant. Rearranging:

    $P = \left( \frac{nR}{V} \right) T$

    This is of the form $y = mx$, a straight line through the origin. The slope $m$ is $nR/V$. Since $n$ (number of moles) and $R$ (gas constant) are the same for all lines in the graph, the slope depends only on $V$: a larger volume gives a smaller slope, and a smaller volume gives a larger slope.

    Now let’s apply this to the given graph.

    1. Identify the slopes from the graph

       The four lines are labelled $V_1, V_2, V_3, V_4$ in order of decreasing steepness. That is, $V_1$ has the steepest slope, $V_2$ is next, then $V_3$, and $V_4$ has the shallowest slope.

       So: slope$_1$ > slope$_2$ > slope$_3$ > slope$_4$.

    2. Relate slope to volume

       From $P = (nR/V) T$, the slope is $nR/V$. Since $nR$ is constant, slope is inversely proportional to $V$:

    $\text{slope} \propto \frac{1}{V}$

    Therefore, a larger slope means a smaller volume, and a smaller slope means a larger volume.

    3. Order the volumes Because slope$_1$ > slope$_2$ > slope$_3$ > slope$_4$, we have:

    $\frac{1}{V_1} > \frac{1}{V_2} > \frac{1}{V_3} > \frac{1}{V_4}$

    Taking reciprocals (which reverses the inequality order):

    $V_1 < V_2 < V_3 < V_4$

    4. Match with the options The correct relation is $V_1 < V_2 < V_3 < V_4$, which corresponds to option (D).

    Watch out

    A common mistake is to think that a steeper line means a larger volume, because “more pressure” might feel like “more gas”. But remember: for a fixed amount of gas, a steeper P–T line means the gas is confined to a smaller volume — it takes less temperature increase to raise the pressure sharply.

    Tip

    If you ever forget the direction, test with a simple case: imagine a very small volume (like a rigid tank). A small temperature increase will cause a huge pressure jump — steep slope. A huge volume (like a balloon) gives a gentle slope. So steep slope = small volume.

    ✓Final answer

    The correct option is (D).

    ANSWER: D