Class 11 Physics - Thermodynamics (Important Q&A)
Thermodynamics Marks · 2M 3M
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📘 2 Mark Q&A
3+ Q&ASolution:
The First Law of Thermodynamics states that heat supplied to a system is partly used to increase its internal energy and partly used to do work.
The mathematical form is
\( \delta Q = dU + \delta W \).
For finite changes,
\( Q = \Delta U + W \).
Where:
\( Q \) = Heat supplied to the system.
\( \Delta U \) = Change in internal energy of the system.
\( W \) = Work done by the system.
If the work done is due to change in volume, then
\( \delta W = P\,dV \).
So the equation becomes \( \delta Q = dU + P\,dV \).
Solution:
Reversible Process: A reversible process is a thermodynamic process that can be reversed in such a way that both the system and surroundings are restored exactly to their initial states, without any net change.
Example: Slow isothermal expansion or compression of an ideal gas by applying or removing an infinitesimally small external pressure.
Irreversible Process: An irreversible process is a process that cannot be reversed to bring both the system and surroundings back to their original states.
Example: Free expansion of a gas into vacuum or heat flow from a hot body to a cold body.
Solution:
Intensive Variable: An intensive variable is a physical quantity that does not depend on the amount of substance present in the system.
Examples: Temperature \(T\), Pressure \(P\), Density \( \rho \).
Extensive Variable: An extensive variable is a physical quantity that depends on the amount of substance present in the system.
Examples: Volume \(V\), Mass \(m\), Internal Energy \(U\).
📗 3 Mark Q&A
5+ Q&ASolution:
For an ideal gas, the First Law of Thermodynamics is
\( \delta Q = dU + P\,dV \)
At constant volume, \( dV = 0 \)
So, \( \delta Q_v = dU \)
By definition, \( C_v = \left( \dfrac{\delta Q}{dT} \right)_V \)
\( \Rightarrow \delta Q_v = C_v \, dT \)
Hence, \( dU = C_v \, dT \)
At constant pressure,
\( \delta Q_p = dU + P\,dV \)
Substituting \( dU = C_v dT \), we get
\( \delta Q_p = C_v dT + P\,dV \, \, \, .....(i)\)
From ideal gas equation \( PV = RT \) (for one mole)
Differentiating, \( P\,dV + V\,dP = R\,dT \)
At constant pressure, \( dP = 0 \)
So, \( P\,dV = R\,dT \)
Therefore, equation \((i)\) becomes \( \delta Q_p = C_v dT + R dT \)
\( \delta Q_p = (C_v + R)\, dT \, \, \, .....(ii) \)
By definition, \( C_p = \left( \dfrac{\delta Q}{dT} \right)_P \)
\( \Rightarrow \delta Q_p = C_p \, dT \)
From equation \( (ii) \), we get
\( \Rightarrow (C_v + R)\, dT = C_p \, dT \)
Hence, \( C_p = C_v + R \)
\( C_p - C_v = R \).
Hence proved.
Solution:
For an isothermal process of an ideal gas, \( PV = \text{constant} \)
Differentiating, \( P\,dV + V\,dP = 0 \)
\( \Rightarrow V\,dP = -P\,dV \)
\( \therefore \left( \dfrac{dP}{dV} \right)_{\text{iso}} = -\dfrac{P}{V} \)
For an adiabatic process, \( PV^{\gamma} = \text{constant} \)
Differentiating, \( P\gamma V^{\gamma-1} dV + V^{\gamma} dP = 0 \)
\( \Rightarrow \gamma \dfrac{P}{V} dV + dP = 0 \)
\( \therefore \left( \dfrac{dP}{dV} \right)_{\text{adi}} = -\gamma \dfrac{P}{V} \)
Since \( \gamma > 1 \),
\( \Rightarrow \left| \dfrac{dP}{dV} \right|_{\text{adi}} = \gamma \dfrac{P}{V} > \dfrac{P}{V} = \left| \dfrac{dP}{dV} \right|_{\text{iso}} \)
Therefore, the slope of the adiabatic curve is greater than that of the isothermal curve.
Hence, adiabatic curves are steeper than isothermal curves.
Solution:
For an adiabatic process, \( PV^{\gamma} = \text{constant} \).
For one mole of an ideal gas, \( PV = RT \).
So, \( P = \dfrac{RT}{V} \).
On substituting in \( PV^{\gamma} = \text{constant} \), we get
\( \dfrac{RT}{V} \cdot V^{\gamma} = \text{constant} \).
\( \Rightarrow RT \, V^{\gamma - 1} = \text{constant} \).
Since \(R\) is constant,
\( \therefore T V^{\gamma - 1} = \text{constant} \).
Where \( \gamma = \dfrac{C_p}{C_v} \).
Hence proved.
Solution:
Heat Engine: A heat engine is a device which converts heat energy into mechanical work by operating in a cyclic process between a high temperature reservoir and a low temperature reservoir.
It absorbs heat \(Q_H\) from a hot source, performs work \(W\), and rejects heat \(Q_C\) to a cold sink.
Adiabatic and Isothermal Curves on \(P\text{-}V\) Diagram:
The adiabatic curve is steeper than the isothermal curve because \( \left( \dfrac{dP}{dV} \right)_{\text{adi}} = -\gamma \dfrac{P}{V} \) and \( \left( \dfrac{dP}{dV} \right)_{\text{iso}} = -\dfrac{P}{V} \), where \( \gamma > 1 \).
Thus, for the same decrease in volume, pressure increases more rapidly in the adiabatic process than in the isothermal process.
Solution:
Given:
Heat absorbed from source \( Q_H = 200 \, \text{cal} \)
Heat rejected to sink \( Q_C = 150 \, \text{cal} \)
Temperature of source \( T_H = 500 \, \text{K} \)
For a Carnot engine, \( \dfrac{Q_C}{Q_H} = \dfrac{T_C}{T_H} \)
So, \( \dfrac{150}{200} = \dfrac{T_C}{500} \)
\( \Rightarrow \dfrac{3}{4} = \dfrac{T_C}{500} \)
\( \Rightarrow T_C = \dfrac{3}{4} \times 500 \)
\( \therefore T_C = 375 \, \text{K} \)
Efficiency of Carnot engine is
\( \Rightarrow \eta = 1 - \dfrac{T_C}{T_H} \)
\( \Rightarrow \eta = 1 - \dfrac{375}{500} \)
\( \Rightarrow \eta = 1 - 0.75 \)
\( \therefore \eta = 0.25 \)
So, efficiency \( = 25\% \)
Answer: The temperature of sink is \( 375 \, \text{K} \) and efficiency is \( 25\% \).
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