Thermodynamics
11.1 – 11.2 Specific Heat & Calorimetry
Q11.1 Geyser Fuel Consumption.
Given: Flow rate = $3.0 \text{ L/min}$. Mass rate $m = 3000 \text{ g/min}$.
Temp rise $\Delta T = 77 – 27 = 50^\circ \text{C}$.
Heat of combustion $H = 4.0 \times 10^4 \text{ J/g}$.
Specific heat of water $c = 4.2 \text{ J g}^{-1} \text{C}^{-1}$.
Heat required per minute:
$Q = mc\Delta T = 3000 \times 4.2 \times 50 = 630,000 \text{ J/min}$.
Rate = $\frac{630000}{40000} = \frac{63}{4} = \mathbf{15.75 \text{ g/min}}$.
Q11.2 Heating Nitrogen Gas ($C_p$).
Given: Mass $m = 2.0 \times 10^{-2} \text{ kg} = 20 \text{ g}$.
Molar mass $M = 28 \text{ g/mol}$. $\Delta T = 45^\circ \text{C}$.
Number of moles $n = \frac{m}{M} = \frac{20}{28} \approx 0.714$.
Nitrogen ($N_2$) is a diatomic gas. At room temperature:
$C_v = \frac{5}{2}R$, and $C_p = C_v + R = \frac{7}{2}R$.
$C_p = 3.5 \times 8.3 = 29.05 \text{ J mol}^{-1} \text{K}^{-1}$.
Heat supplied at constant pressure:
$Q = n C_p \Delta T = \frac{20}{28} \times 29.05 \times 45$.
$Q = 0.714 \times 1307.25 \approx \mathbf{933.4 \text{ J}}$.
11.3 Conceptual Thermodynamics
Q11.3 Reasoning (Temp Mix, Coolants, Tyres, Climate).
(a) Final Temperature Mean:
Heat lost = Heat gained. $m_1 c_1 (T_1 – T) = m_2 c_2 (T – T_2)$.
The final temperature $T$ depends on mass ($m$) and specific heat ($c$). It is only the mean $(T_1+T_2)/2$ if $m_1 c_1 = m_2 c_2$.
(b) Coolants high specific heat:
A coolant needs to absorb a large amount of heat ($Q$) from the reactor with a minimal rise in its own temperature ($\Delta T$). High specific heat ($c$) ensures large $Q$ for small $\Delta T$.
(c) Tyre pressure increases:
During driving, friction and flexing work on the tyre generate heat. The temperature ($T$) of the air inside increases. For constant volume ($V$), Pressure $P \propto T$ (Gay-Lussac’s Law). Hence pressure rises.
(d) Harbour town climate:
Water has a high specific heat capacity compared to sand/land. The ocean absorbs heat slowly during the day (keeping air cool) and releases it slowly at night (keeping air warm), moderating the climate.
11.4 – 11.6 Processes & First Law
Q11.4 Adiabatic Compression of Hydrogen.
Condition: Insulated walls + sand (rapid/insulated) = Adiabatic Process.
Gas: Hydrogen (Diatomic). $\gamma = 1.4$.
Volume change: $V_2 = \frac{V_1}{2}$.
Adiabatic Relation: $P_1 V_1^\gamma = P_2 V_2^\gamma$.
$\frac{P_2}{P_1} = \left(\frac{V_1}{V_2}\right)^\gamma = (2)^{1.4}$.
Using log or calculator: $2^{1.4} \approx 2.639$.
Pressure increases by a factor of 2.64.
Q11.5 First Law Application (Path Dependence).
Path 1 (Adiabatic):
$Q = 0$. Work done on system $= 22.3 \text{ J}$. So work by system $W_1 = -22.3 \text{ J}$.
First Law: $\Delta U = Q – W = 0 – (-22.3) = +22.3 \text{ J}$.
Since $\Delta U$ is a state function, it is the same for Path 2.
Path 2:
Heat absorbed $Q_2 = 9.35 \text{ cal} = 9.35 \times 4.19 \approx 39.18 \text{ J}$.
$\Delta U = Q_2 – W_2 \Rightarrow 22.3 = 39.18 – W_2$.
$W_2 = 39.18 – 22.3 = \mathbf{16.88 \text{ J}}$.
(Work done by the system).
Q11.6 Free Expansion.
System thermally insulated ($Q=0$). Expansion into vacuum ($P_{ext}=0 \Rightarrow W=0$).
From First Law: $\Delta U = Q – W = 0$.
(a) Final Pressure:
For ideal gas, if $\Delta U = 0$, then $\Delta T = 0$ (Isothermal).
Initial: $P_1, V$. Final: $P_2, 2V$.
Boyle’s Law: $P_1 V = P_2 (2V) \Rightarrow P_2 = P_1/2$.
At STP, $P_1 = 1 \text{ atm}$. Final pressure = 0.5 atm.
(b) Change in Internal Energy: Zero.
(c) Change in Temperature: Zero.
(d) Intermediate States: No. Free expansion is a rapid, non-equilibrium process. State variables ($P, T$) are not well-defined for the whole system during the process.
11.7 – 11.8 Rates & Work Calculation
Q11.7 Rate of Internal Energy Increase.
First Law in rate form: $\frac{dU}{dt} = \frac{dQ}{dt} – \frac{dW}{dt}$.
Heat rate $\frac{dQ}{dt} = 100 \text{ W}$ (Supplied $\rightarrow +$).
Work rate $\frac{dW}{dt} = 75 \text{ W}$ (Performed by system $\rightarrow +$).
Q11.8 Work Done in Cycle D-E-F.
Note: Assuming standard NCERT graph values where D is at higher volume/pressure and F closes the loop. Let’s interpret based on typical description. Usually D to E is expansion, E to F is compression.
Work Done = Area under the P-V curve.
Path D to E (Linear):
Area is a Trapezoid under line DE.
$W_{DE} = \text{Area of Trapezium} = \frac{1}{2}(P_D + P_E)(V_E – V_D)$.
Path E to F (Isobaric Compression):
Pressure is constant ($P_E$ or $P_F$). Volume reduces.
$W_{EF} = P (V_F – V_E) = -P \Delta V$.
Total Work: $W_{total} = W_{DE} + W_{EF}$.
(Requires reading numeric values of P and V from the provided figure in the textbook).
Example Calculation (if D=(2, 600) and E=(5, 300) and F=(2, 300)):
$W_{DE} = \frac{1}{2}(600 + 300)(5 – 2) = 450 \times 3 = 1350 \text{ J}$.
$W_{EF} = 300(2 – 5) = 300(-3) = -900 \text{ J}$.
$W_{total} = 1350 – 900 = \mathbf{450 \text{ J}}$.
(This represents the area of triangle DEF).