Thermodynamics Chapter-Wise Test 2

In an isochoric process, volume remains constant (\( \Delta V = 0 \)). Since work done is \( W = P \Delta V \), no work is done by the gas (\( W = 0 \)), and all heat supplied changes the internal energy.

First Law: \( \Delta Q = \Delta U + \Delta W \).

\( \Delta Q = 700 \), \( \Delta W = -300 \) (work done on system, negative work by system).

\( 700 = \Delta U - 300 \Rightarrow \Delta U = 700 + 300 = 1000 \, \text{J} \).

The Zeroth Law establishes that two systems in thermal equilibrium with a third have the same temperature, providing the basis for a consistent temperature scale and thermometer calibration.

The Kelvin-Planck statement of the Second Law prohibits a heat engine from converting all heat from a single reservoir into work without rejecting some to a colder reservoir, as this would violate the natural tendency toward equilibrium.

\( \Delta Q = m L \).

\( m = 1.4 \), \( L = 2256 \).

\( \Delta Q = 1.4 \times 2256 = 3158.4 \, \text{J} \approx 3158 \, \text{J} \).

\( W = P \Delta V \), \( P V = \mu R T \).

\( \Delta V = 12 - 6 = 6 \, \text{L} \).

Initial \( P = \frac{\mu R T}{V_1} \), but directly: \( W = \mu R T \left(\frac{V_2 - V_1}{V_1}\right) \), adjust via \( P \Delta V \).

\( W = 0.8 \times 8.3 \times 350 \times \frac{6}{6} = 2324 \, \text{J} \) (corrected: \( W = \mu R T \)).

No, \( W = P \Delta V \), use \( \mu R T \) correctly: \( W = 0.8 \times 8.3 \times 350 = 2324 \, \text{J} \) (unit consistency).

\( \Delta U = m s \Delta T \) (no work in constant volume or solid).

\( m = 0.1 \), \( s = 127.7 \), \( \Delta T = 40 - 20 = 20 \).

\( \Delta U = 0.1 \times 127.7 \times 20 = 255.4 \, \text{J} \).

The First Law (\( \Delta Q = \Delta U + \Delta W \)) is a conservation principle, stating heat added equals internal energy increase plus work done. Option D is correct.

\( P_1 V_1^\gamma = P_2 V_2^\gamma \).

\( P_1 = 1 \, \text{atm} \), \( V_1 = 8 \, \text{L} \), \( V_2 = 2 \, \text{L} \), \( \gamma = 1.4 \).

\( 1 \times 8^{1.4} = P_2 \times 2^{1.4} \).

\( P_2 = \frac{8^{1.4}}{2^{1.4}} = \left(\frac{8}{2}\right)^{1.4} = 4^{1.4} \).

\( 4^{1.4} = (2^2)^{1.4} = 2^{2.8} \approx 6.96 \, \text{atm} \).

For isothermal: \( W = \mu R T \ln\left(\frac{V_2}{V_1}\right) \).

\( \mu = 0.4 \), \( R = 8.3 \), \( T = 420 \), \( V_2 = 21 \), \( V_1 = 7 \).

\( W = 0.4 \times 8.3 \times 420 \times \ln\left(\frac{21}{7}\right) = 1394.4 \times \ln(3) \).

\( \ln(3) \approx 1.0986 \), \( W \approx 1394.4 \times 1.0986 \approx 1532 \, \text{J} \).

Specific heat capacity varies with temperature because the energy required to raise the temperature of a substance depends on molecular interactions and vibrational modes, which change with temperature (e.g., water’s variation in Fig. 11.5).

The law of equipartition predicts that each atom in a solid has 3 degrees of freedom (vibrational), contributing \( \frac{3}{2} k_B T \) kinetic and potential energy per atom. For a mole, \( U = 3 R T \), so \( C = \frac{\Delta U}{\Delta T} = 3R \), which matches experimental values at ordinary temperatures.

For an ideal gas in an isothermal process, \( \Delta U = 0 \) (since \( U \) depends only on temperature).

\( \Delta Q = \mu C_p \Delta T \).

\( \mu = 0.8 \), \( C_p = 29.1 \), \( \Delta T = 410 - 330 = 80 \).

\( \Delta Q = 0.8 \times 29.1 \times 80 = 1862.4 \, \text{J} \approx 1862 \, \text{J} \).

The First Law (\( \Delta Q = \Delta U + \Delta W \)) is a conservation of energy principle, not requiring equilibrium or constant temperature. Option C is incorrect as it imposes an unnecessary condition.