IB Physics Practice Exam — Free

Apply the first law of thermodynamics, $Q = \Delta U + W$, where $W$ is the work done by the gas.

The container is rigid, so the volume does not change and the gas does no work: $W = p\,\Delta V = 0$. All of the thermal energy supplied therefore goes into internal energy, $Q = \Delta U$.

For an ideal monatomic gas the internal energy is entirely the translational kinetic energy of the molecules, $U = \tfrac{3}{2}nRT$, so

The initial pressure is not needed for the energy calculation; it would only be needed to find the final pressure, which rises to $1.2\times 10^{5}(400/300) = 1.6 \times 10^{5}$ Pa because $p/T$ is constant at fixed volume.

The tempting answer 0 J confuses zero work with zero heat: the gas does no work, but it certainly gains energy, as the temperature rise shows. 42 J is $nR\,\Delta T$, which is the work done in a constant-pressure expansion, not the change in internal energy; 104 J is $\tfrac{5}{2}nR\,\Delta T$, the heat needed at constant pressure, where the gas would also expand and do work.

$Q = 62\,\text{J}$

(a) During the collision the only significant forces are the ones the cars exert on each other, which are internal to the two-car system, so total momentum is conserved. Take the direction of the moving car as positive:

The combined mass moves in the original direction at 9.0 m s⁻¹, slower than 15 m s⁻¹ because the same momentum is now shared by more mass.

(b) Kinetic energy before:

Kinetic energy after:

Forty percent of the original kinetic energy is lost, which is exactly what a perfectly inelastic collision between these masses must give: the fraction retained is $m_1/(m_1 + m_2) = 0.6$. The lost energy is not destroyed; it is transferred to internal energy (the cars warm up), to sound, and to the work done permanently deforming the bodywork. Momentum is conserved while kinetic energy is not, which is the signature of an inelastic collision.

(a) $v = 9.0\,\text{m s}^{-1}$; (b) 54 kJ lost, converted to internal (thermal) energy, sound and deformation of the cars

(a) Combine the parallel pair first:

The internal resistance is in series with this, so the total resistance of the circuit is $R_p + r = 2.0 + 0.50 = 2.5\,\Omega$. The emf drives the current around the whole loop:

(b) Part of the emf is used driving the current through the internal resistance (the 'lost volts'), so the terminal potential difference is

This is also the potential difference across each parallel resistor, since $V = IR_p = 4.8(2.0) = 9.6$ V.

(c) The 3.0 Ω resistor has 9.6 V across it, so

Equivalently its current is $I_3 = 9.6/3.0 = 3.2$ A and $P = I^2R = 3.2^2(3.0) = 30.7$ W. The 6.0 Ω resistor carries the other 1.6 A and dissipates 15.4 W.

(d) Total power generated by the cell is $P_{total} = \varepsilon I = 12(4.8) = 57.6$ W. Power delivered to the external circuit is $VI = 9.6(4.8) = 46.1$ W (which equals 30.7 W + 15.4 W). The fraction is

The remaining 20% (11.5 W) is dissipated as heat inside the cell. The efficiency equals $R_p/(R_p + r)$, so it improves as the external resistance becomes large compared with the internal resistance.

(a) $I = 4.8\,\text{A}$; (b) $V = 9.6\,\text{V}$; (c) $P \approx 31\,\text{W}$; (d) 0.80 (80%)