NCERT Solutions · Class 11 Physics Thermodynamics

A geyser is a heater, so "rate of fuel consumption" is just heat needed per minute divided by heat released per gram of fuel.Two energy quantities are being equated: the heat that must be delivered to the water every minute, and the heat that burning fuel supplies. Setting them equal gives the mass of fuel burned per minute.Step $\displaystyle 1$: Convert the water flow to a mass flow.The density of water is \(\displaystyle 1000\ \text{kg/m}^3\), which is the same as saying $\displaystyle 1$ litre of water has a mass of $\displaystyle 1$ kg. (This is the fact people forget — litres measure volume, not mass; the "$\displaystyle 1$ L = $\displaystyle 1$ kg" shortcut works only for water, at ordinary temperature, because its density is so close to $\displaystyle 1000$ kg/m³.)\[\dot m = 3.0\ \text{L/min} \times 1\ \text{kg/L} = 3.0\ \text{kg/min} \]Step $\displaystyle 2$: Find the temperature rise.\[\Delta T = 77\,^\circ\text{C} - 27\,^\circ\text{C} = 50\,^\circ\text{C} = 50\ \text{K} \]A difference of temperature is the same number in °C and in K, since the two scales differ only by an additive constant that cancels out.Step $\displaystyle 3$: Apply the specific heat capacity relation.The heat needed to raise a mass \(\displaystyle m\) of water through a temperature change \(\displaystyle \Delta T\) is\[Q = m\,c\,\Delta T \]where \(\displaystyle c\) is the specific heat capacity of water, \(\displaystyle c = 4186\ \text{J kg}^{-1}\text{K}^{-1}\) (the standard value tabulated for water). Using the mass flow rate found above, this becomes a rate of heat delivery:\[\dot Q = \dot m\,c\,\Delta T = (3.0\ \text{kg/min})(4186\ \text{J kg}^{-1}\text{K}^{-1})(50\ \text{K}) \]\[\dot Q = 6.279\times10^{5}\ \text{J/min} \]Step $\displaystyle 4$: Convert heat needed into fuel burned, using the heat of combustion.The heat of combustion, \(\displaystyle 4.0\times10^{4}\ \text{J/g}\), tells us how much heat comes out of burning one gram of fuel. So the rate at which fuel must be burned is the rate of heat demand divided by heat released per gram:\[\dot m_{\text{fuel}} = \frac{\dot Q}{\text{heat of combustion}} = \frac{6.279\times10^{5}\ \text{J/min}}{4.0\times10^{4}\ \text{J/g}} \]\[\dot m_{\text{fuel}} = 15.7\ \text{g/min (unrounded)} \]Step $\displaystyle 5$: Round to the correct number of significant figures.Every measured input in this problem — the flow rate \(\displaystyle 3.0\) L/min, and the heat of combustion \(\displaystyle 4.0\times10^{4}\) J/g — is given to $\displaystyle 2$ significant figures, so the final answer can carry no more precision than that. Rounding \(\displaystyle 15.6975\) to $\displaystyle 2$ significant figures gives \(\displaystyle 16\) g/min.Answer: The fuel must be burned at a rate of about \(\displaystyle 16\ \text{g/min}\) (≈ \(\displaystyle 15.7\) g/min before rounding).

"At constant pressure" is the whole problem — it tells you to use \(\displaystyle C_p\), not \(\displaystyle C_v\). Nitrogen can absorb heat either at constant volume or constant pressure, and it takes more heat to raise its temperature at constant pressure because some of that heat does work pushing back the surroundings as the gas expands. Using \(\displaystyle C_v\) here is the mistake to avoid.Step $\displaystyle 1$: Convert the given mass to moles.Number of moles, \[n = \frac{\text{mass}}{\text{molar mass}} = \frac{2.0\times10^{-2}\ \text{kg}}{28\times10^{-3}\ \text{kg mol}^{-1}} = \frac{20\ \text{g}}{28\ \text{g mol}^{-1}} = \frac{5}{7}\ \text{mol} \approx 0.714\ \text{mol} \]Step $\displaystyle 2$: Find the molar specific heat at constant pressure, \(\displaystyle C_p\).Nitrogen, \(\displaystyle N_2\), is a diatomic molecule. At room temperature it has $\displaystyle 3$ translational and $\displaystyle 2$ rotational degrees of freedom (the vibrational mode is not excited at room temperature), so its molar specific heat at constant volume is \[C_v = \frac{5}{2}R \] Mayer's relation connects \(\displaystyle C_p\) and \(\displaystyle C_v\) for an ideal gas: \[C_p - C_v = R \quad\Rightarrow\quad C_p = C_v + R = \frac{5}{2}R + R = \frac{7}{2}R \] With \(\displaystyle R = 8.3\ \text{J mol}^{-1}\text{K}^{-1}\), \[C_p = \frac{7}{2}\times 8.3\ \text{J mol}^{-1}\text{K}^{-1} = 29.05\ \text{J mol}^{-1}\text{K}^{-1} \]Step $\displaystyle 3$: Apply the heat-capacity relation at constant pressure.The heat needed to change temperature by \(\displaystyle \Delta T\) at constant pressure is \[\Delta Q = n\,C_p\,\Delta T \] where \(\displaystyle n\) is the number of moles, \(\displaystyle C_p\) is the molar specific heat at constant pressure, and \(\displaystyle \Delta T\) is the temperature rise.Substituting \(\displaystyle n = \dfrac{5}{7}\ \text{mol}\), \(\displaystyle C_p = \dfrac{7}{2}R\), and \(\displaystyle \Delta T = 45\ \text{°C} = 45\ \text{K}\) (a change in temperature is the same number of kelvin as of degrees Celsius, so no offset is needed here):\[\Delta Q = \frac{5}{7}\ \text{mol} \times \frac{7}{2}\times 8.3\ \text{J mol}^{-1}\text{K}^{-1} \times 45\ \text{K} \]The factor of \(\displaystyle 7\) cancels neatly: \[\Delta Q = \frac{5}{2}\times 8.3\ \text{J K}^{-1} \times 45\ \text{K} = 20.75\ \text{J K}^{-1} \times 45\ \text{K} = 933.75\ \text{J} \]Step $\displaystyle 4$: Round to the correct number of significant figures.Every measured quantity in the problem — the mass \(\displaystyle 2.0\times10^{-2}\ \text{kg}\), the temperature rise \(\displaystyle 45\ \text{°C}\), and \(\displaystyle R = 8.3\ \text{J mol}^{-1}\text{K}^{-1}\) — is given to $\displaystyle 2$ significant figures, so the final answer can only be trusted to $\displaystyle 2$ significant figures. Rounding \(\displaystyle 933.75\ \text{J}\) accordingly gives \(\displaystyle 9.3\times10^{2}\ \text{J}\).Answer: \(\displaystyle \Delta Q \approx 9.3\times10^{2}\ \text{J}\) (about $\displaystyle 930$ J) of heat must be supplied.

The insulation means no heat can enter or leave — this is an adiabatic process, so you cannot use the ordinary gas law \(\displaystyle PV = \text{constant}\); you need the adiabatic relation instead.Because the cylinder walls are heat-insulating and the piston is also insulated (by the sand), no heat is exchanged between the gas and its surroundings during the compression: \[Q = 0 \] This is the defining condition of an adiabatic process. For an ideal gas undergoing a quasi-static adiabatic change, pressure and volume are related not by \(\displaystyle PV = \text{constant}\) (that only holds at constant temperature) but by\[PV^{\gamma} = \text{constant} \]where \(\displaystyle \gamma = C_p/C_v\) is the ratio of specific heats of the gas.Why \(\displaystyle \gamma = 7/5\) here. Hydrogen, \(\displaystyle \mathrm{H_2}\), is a diatomic gas. A diatomic molecule has $\displaystyle 5$ degrees of freedom ($\displaystyle 3$ translational + $\displaystyle 2$ rotational) at ordinary temperatures, so \[C_v = \frac{5}{2}R, \qquad C_p = C_v + R = \frac{7}{2}R \] \[\gamma = \frac{C_p}{C_v} = \frac{7/2\,R}{5/2\,R} = \frac{7}{5} = 1.4 \] (The number of moles, $\displaystyle 3$, and the fact that the gas starts at standard temperature and pressure don't enter the calculation at all — they only fix the starting point, not the ratio by which pressure changes. Don't be tempted to bring \(\displaystyle n\), \(\displaystyle T\), or \(\displaystyle P_{\text{initial}}\) into the algebra; they cancel.)Setting up the ratio. Let state $\displaystyle 1$ be before compression (volume \(\displaystyle V_1\), pressure \(\displaystyle P_1\)) and state $\displaystyle 2$ be after compression to half the volume: \[V_2 = \frac{V_1}{2} \] Applying \(\displaystyle PV^\gamma = \text{constant}\) to the two states: \[P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \] Solve for the pressure ratio: \[\frac{P_2}{P_1} = \left(\frac{V_1}{V_2}\right)^{\gamma} = \left(\frac{V_1}{V_1/2}\right)^{\gamma} = 2^{\gamma} \]Evaluating the number. With \(\displaystyle \gamma = 1.4\): \[\frac{P_2}{P_1} = 2^{1.4} = 2^{1}\times 2^{0.4} \] Since \(\displaystyle 2^{0.4} = e^{0.4 \ln 2} = e^{0.4 \times 0.6931} = e^{0.2773} \approx 1.3195\), \[\frac{P_2}{P_1} \approx 2 \times 1.3195 = 2.639 \]The volume was halved exactly, and \(\displaystyle \gamma = 7/5\) is an exact value for an ideal diatomic gas, so the limiting precision comes from the exponential evaluation itself — quoting the result to four significant figures is appropriate.Notice that this factor (\(\displaystyle \approx 2.64\)) is larger than the factor of $\displaystyle 2$ you'd get from an isothermal compression (\(\displaystyle P_2/P_1 = V_1/V_2 = 2\) if \(\displaystyle T\) were held constant). That's the physical content of the result: because no heat escapes, all the work done in pushing the piston in goes into raising the gas's internal energy, so the temperature also rises during the compression — and a hotter gas at a given volume exerts extra pressure on top of what compression alone would cause.Answer: The pressure increases by a factor of \(\displaystyle 2^{7/5} \approx 2.639\) (about $\displaystyle 2.64$ times the original pressure).

Internal energy depends only on the initial and final states, not on the path — so the same \(\displaystyle U_B - U_A \) applies to both processes.The first law of thermodynamics states \[\Delta Q = \Delta U + \Delta W \] where \(\displaystyle \Delta Q \) is the heat absorbed by the system, \(\displaystyle \Delta U \) is the change in internal energy, and \(\displaystyle \Delta W \) is the work done by the system.Step $\displaystyle 1$: Find \(\displaystyle \Delta U \) from the adiabatic path.In an adiabatic process no heat is exchanged, so \(\displaystyle \Delta Q = 0 \).The problem says $\displaystyle 22.3$ J of work is done on the system, which means the work done by the system is negative: \[\Delta W_1 = -22.3\ \text{J} \]Substituting into the first law for path $\displaystyle 1$ (A → B, adiabatic): \[0 = \Delta U + (-22.3\ \text{J}) \] \[\Delta U = 22.3\ \text{J} \]This is the change in internal energy of the gas between state A and state B. Because internal energy is a state function, this value is the same no matter which route is taken from A to B — this is exactly the property that lets the second (non-adiabatic) path be solved without knowing any details of that process.Step $\displaystyle 2$: Convert the heat given for the second path into joules.The heat absorbed in the second process is given in calories, but the first law needs consistent units: \[\Delta Q_2 = 9.35\ \text{cal} \times 4.19\ \dfrac{\text{J}}{\text{cal}} = 39.1765\ \text{J} \]Step $\displaystyle 3$: Apply the first law to the second path.\[\Delta Q_2 = \Delta U + \Delta W_2 \] \[39.1765\ \text{J} = 22.3\ \text{J} + \Delta W_2 \] \[\Delta W_2 = 39.1765\ \text{J} - 22.3\ \text{J} = 16.8765\ \text{J} \]Step $\displaystyle 4$: Round to the correct number of significant figures.The least precise inputs ($\displaystyle 9.35$ cal, $\displaystyle 4.19$ J/cal, $\displaystyle 22.3$ J) all carry $\displaystyle 3$ significant figures, so the result is rounded to $\displaystyle 3$ significant figures at the very end, not at any intermediate step: \[\Delta W_2 \approx 16.9\ \text{J} \]Since \(\displaystyle \Delta W_2 \) is positive, the system does this work outward (work done by the gas), unlike the first path where work was done on the gas.Answer: The net work done by the system in the second process is $\displaystyle 16.9$ J (done by the gas on the surroundings).

Free expansion into vacuum does no work and exchanges no heat, so the internal energy — and hence the temperature — of the gas does not change at all.(a) Final pressureCylinder A holds the gas at standard temperature and pressure, \(\displaystyle P_1 = 1\ \text{atm} \), filling its own volume \(\displaystyle V \) (its capacity). Cylinder B has the same capacity \(\displaystyle V \) and is empty.When the stopcock is opened, the gas is free to occupy both cylinders, so the final volume is \[V_2 = V_A + V_B = V + V = 2V. \]This is a free expansion (Joule expansion): the gas expands into a vacuum, so there is nothing pushing back on it. Work done by a gas against its surroundings is \(\displaystyle W = \int P_{\text{ext}}\, dV \), and here \(\displaystyle P_{\text{ext}} = 0 \) (empty space), so \(\displaystyle W = 0 \). The whole apparatus is thermally insulated, so no heat enters or leaves either: \(\displaystyle Q = 0 \).By the first law of thermodynamics, \[\Delta U = Q - W = 0 - 0 = 0. \] For an ideal gas, internal energy depends only on temperature, so \(\displaystyle \Delta U = 0 \) forces \(\displaystyle \Delta T = 0 \). (This is the step people get wrong: "isothermal" here does not mean heat was added slowly to keep \(\displaystyle T \) fixed — no heat moved at all. The temperature simply never changes because there is no mechanism, work or heat, to change it.)With \(\displaystyle T \) and the number of moles \(\displaystyle n \) both unchanged, the ideal gas equation \(\displaystyle PV = nRT \) gives, at the two states, \[P_1 V_1 = P_2 V_2 \] \[P_2 = P_1 \frac{V_1}{V_2} = (1\ \text{atm}) \times \frac{V}{2V} = 0.5\ \text{atm}. \]Since the stopcock stays open, A and B form one connected vessel with a single uniform pressure throughout — $\displaystyle 0.5$ atm in A and $\displaystyle 0.5$ atm in B, not two different values.(b) Change in internal energy and change in temperatureBoth follow directly from the working above: because the gas does no work and exchanges no heat during free expansion, \(\displaystyle \Delta U = 0 \). Because the internal energy of an ideal gas is a function of temperature alone, \(\displaystyle \Delta U = 0 \) means \(\displaystyle \Delta T = 0 \) — the gas is back at standard temperature once it settles down, exactly where it started.It is tempting to expect a gas to cool when its volume increases, but that cooling only happens when the gas does work in expanding (for example, pushing a piston outward). Here there is no piston and no resistance to push against, so no work is done and, for an ideal gas, no temperature change occurs.(d) Do the intermediate states lie on the P-V-T surface?No. The P-V-T surface is a plot of equilibrium states — states in which pressure, volume, and temperature are each single, well-defined, uniform numbers for the whole gas. Free expansion is sudden and uncontrolled: the instant the stopcock opens, gas rushes into B, and for some time afterward the density, pressure, and local temperature are not the same everywhere in the connected cylinders — they have not yet had time to equalize. Such an intermediate state has no single value of \(\displaystyle P \), \(\displaystyle V \), or \(\displaystyle T \) that describes the whole system, so it cannot be plotted as a point on the P-V-T surface at all. Only the initial state (gas confined to A, before the stopcock opens) and the final settled state (gas spread uniformly through A and B) are true equilibrium states lying on the surface; everything the system passes through in between does not.Answer: (a) The final pressure is $\displaystyle 0.5$ atm in both A and B. (b) \(\displaystyle \Delta U = 0 \) and \(\displaystyle \Delta T = 0 \). (d) No — the intermediate states are non-equilibrium states with no uniform P, V, T for the system, so they do not lie on the P-V-T surface.

This is the first law of thermodynamics applied to rates instead of totals — heat in splits between raising internal energy and doing work.The first law of thermodynamics states \[\Delta Q = \Delta U + \Delta W \] where \(\displaystyle \Delta Q \) is the heat supplied to the system, \(\displaystyle \Delta U \) is the increase in the system's internal energy, and \(\displaystyle \Delta W \) is the work done by the system on its surroundings. The law is a statement about energy conservation, and it holds equally well when every term is expressed as a rate (energy per unit time) rather than a total amount, since dividing right through by a time interval keeps the equation true: \[\frac{\Delta Q}{\Delta t} = \frac{\Delta U}{\Delta t} + \frac{\Delta W}{\Delta t} \]Here the heater supplies heat at a rate \[\frac{\Delta Q}{\Delta t} = 100\ \text{W} = 100\ \text{J/s} \] and the system does work on its surroundings at a rate \[\frac{\Delta W}{\Delta t} = 75\ \text{J/s} \]A common slip here is to add these two rates instead of subtracting — remember that the heat supplied is the source, and it is shared between raising the internal energy and doing work, so the internal-energy rate is what is left over after the work is paid for: \[\frac{\Delta U}{\Delta t} = \frac{\Delta Q}{\Delta t} - \frac{\Delta W}{\Delta t} \]Substituting the given values: \[\frac{\Delta U}{\Delta t} = 100\ \text{J/s} - 75\ \text{J/s} = 25\ \text{J/s} \]Since $\displaystyle 1$ J/s = $\displaystyle 1$ W, this rate of increase of internal energy can equally be written as $\displaystyle 25$ W. Both given quantities ($\displaystyle 100$ W, $\displaystyle 75$ J/s) have two significant figures, and $\displaystyle 25$ J/s already carries two significant figures, so no further rounding is needed.Answer: The internal energy increases at a rate of $\displaystyle 25$ J/s ($\displaystyle 25$ W).

Work done by a gas is the area under its p-V curve — and that area gets a plus sign while the gas expands, a minus sign while it is compressed.Read the three states straight off the graph. Pressure is on the y-axis in \(\displaystyle \text{N/m}^2 \), volume on the x-axis in \(\displaystyle \text{m}^3 \):\[D = (V_D,\,p_D) = (2.0\ \text{m}^3,\ 600\ \text{N/m}^2) \] \[E = (V_E,\,p_E) = (5.0\ \text{m}^3,\ 300\ \text{N/m}^2) \] \[F = (V_F,\,p_F) = (2.0\ \text{m}^3,\ 300\ \text{N/m}^2) \]Step $\displaystyle 1$ — work done D → E (the sloping straight-line process).For any process, \(\displaystyle W = \displaystyle\int_{V_i}^{V_f} p\,dV \), which on a p-V diagram is just the area under the curve between the two states. Here that area is a trapezium — a rectangle of height equal to the lower pressure, with a triangle sitting on top of it for the extra pressure at D. This split is the safe way to handle a straight-line (not isobaric, not isothermal) process: never treat it as \(\displaystyle p\Delta V \) with a single pressure value.Rectangle, height \(\displaystyle p_E = 300\ \text{N/m}^2 \), width \(\displaystyle V_E-V_D \): \[W_{\text{rect}} = p_E(V_E-V_D) = 300\times(5.0-2.0) = 300\times3.0 = 900\ \text{J} \]Triangle, base \(\displaystyle V_E-V_D = 3.0\ \text{m}^3 \), height \(\displaystyle p_D-p_E = 600-300 = 300\ \text{N/m}^2 \): \[W_{\text{tri}} = \frac12(V_E-V_D)(p_D-p_E) = \frac12\times3.0\times300 = 450\ \text{J} \]\[W_{DE} = W_{\text{rect}}+W_{\text{tri}} = 900+450 = 1350\ \text{J} \]This is positive because the volume is increasing from D to E — the gas is expanding and doing work on its surroundings.Step $\displaystyle 2$ — work done E → F (isobaric compression).This leg is at constant pressure \(\displaystyle p = 300\ \text{N/m}^2 \), so \(\displaystyle W = p\,\Delta V \): \[W_{EF} = p(V_F-V_E) = 300\times(2.0-5.0) = 300\times(-3.0) = -900\ \text{J} \]The pressure here is the same $\displaystyle 300$ N/m² that appeared in Step $\displaystyle 1$, but the sign is not — this leg is a compression (volume falling from $\displaystyle 5.0$ m³ back to $\displaystyle 2.0$ m³), so the work done by the gas is negative: the surroundings are doing work on the gas, not the other way around. Dropping this sign is the single easiest way to get this problem wrong.Step $\displaystyle 3$ — add the two legs.\[W_{DEF} = W_{DE}+W_{EF} = 1350 + (-900) = 450\ \text{J} \]The two $\displaystyle 300$ N/m² isobaric strips (the rectangle under D–E and the compression E–F) are equal in size and opposite in sign, so they cancel exactly, leaving only the triangular wedge from the sloping part of D–E. That triangle's area, $\displaystyle 450$ J, is the entire net result — a useful check on the arithmetic above.All four readings off the graph ($\displaystyle 2.0$, $\displaystyle 5.0$, $\displaystyle 600$, $\displaystyle 300$) carry two or three significant figures, so the total is quoted the same way, with no further rounding needed since the arithmetic came out exact.Answer: The total work done by the gas from D to E to F is \(\displaystyle +450\ \text{J} \) (net work done by the gas on its surroundings).