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Thermodynamics — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsThermodynamics5 Marks
Question
  1. State first law of thermodynamics. What are its limitations? Why $C_p > C_v$?
  2. An electric heater supplies heat to a system at a rate 100W. If the system performs work at a rate of 75 joules per second at what rate is the internal energy increases.
OR
  1. Derive an expression for the work done during the isothermal expression of x mole of an ideal gas.
  2. A steam engine delivers $5.4 \times 10^8J$ of work per minute and services $3.6 \times 10^9J$ of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?
✓

Answer

  1. According to the first law of thermodynamics, the total heat energy change dQ is the sum of the internal energy change dU and work done dW,
i.e. dQ = dU + dW
Limitations of the first law of thermodynamics:
  1. The first law does not indicate the direction in which the heat change can occur.
  2. The first law does not give any idea about the extent of heat change.
  3. The first law of thermodynamics gives no information about the source of heat, i.e. whether it is a hot or cold body.
The relation between two specific heats of a gas, i.e. $C_p$ and $C_v$ is given by $C_p - C_v = R$ where, R is the molar gas constant and is equal to $8.31J ~mole^{-1}K^{-1}$.
$C_p > C_v$ because a part of the energy supplied in the isobaric process goes to increase the volume of the gas and the remaining increases the temperature.
  1. According to first law of thermodynamics,
dQ = dU + dW
Differentiating w.r.t., time (t)
$\frac{\text{dQ}}{\text{dt}}=\frac{\text{dU}}{\text{dt}}+\frac{\text{dW}}{\text{dt}}$
$100\text{W}=\frac{\text{dU}}{\text{dt}}+75$
$[\because$ Given, system performs work at a rate of 75J/ sec.$]$
$\frac{\text{dU}}{\text{dt}}\rightarrow$ Rate of change of internal energy
$\frac{\text{dU}}{\text{dt}}=100-75=25\text{W}$
Alternate Answer
  1. For a small change in volume, work done is given by dW = PdV.
We know, PV = nRT
For T = Constant
$\text{dW}=\text{nRt}\cdot\frac{\text{dV}}{\text{V}}$
$\therefore\text{P}=\frac{\text{nRT}}{\text{V}}$
Net work done under isothermal condition to change the volume from $Y_i$ to $V_f$​​​​​​​ is,
$\text{W}=\int\limits^{\text{V}_\text{f}}_{\text{V}_\text{i}}\text{dW}$
$=\text{nRT}\int\limits^{\text{V}_\text{f}}_{\text{V}_\text{i}}\frac{\text{dV}}{\text{V}}=\text{nRT}|\log_\text{e}\text{V}|^{\text{V}_\text{f}}_{\text{V}_\text{i}}$
$=\text{nRT }\log_\text{e}\Big(\frac{\text{V}_\text{f}}{\text{V}_\text{i}}\Big)$
$\therefore\text{W}=2.3026\text{ nRT }\log_{10}\Big(\frac{\text{V}_\text{f}}{\text{V}_\text{i}}\Big)$
  1. Work done by a steam energy,
$W = 5.4 \times 10^8J, Q_1 = 3.6 \times 10^9J$
Efficiency of the engine, $\eta=?,\text{Q}_2=?$
$\eta=\frac{\text{W}}{\text{Q}_1}=\frac{5.4\times10^8\text{J}} {3.6\times10^9\text{J}}$
$=\frac{54}{36}\times10^{-1}=\frac{3}{20}=0.15$ or 15%
Heat wasted per minute,
$\text{Q}_2=\text{Q}_1-\text{W}$
$\text{Q}_2=3.6\times10^9\text{J}-5.4\times10^8\text{J}$
$=36\times10^8\text{J}-5.4\times10^8\text{J}$
$=10^8(36-5.4)\text{J}=30.6\times10^9\text{J}$

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