IB Physics HL B.4 Thermodynamics Question Bank

Work is done by the gas because there is an increase in volume / gas expands.
So W is positive.
Delta U is greater than zero because P is constant and V increases.
From the first law, Q = Delta U + W, so Q is positive, which means energy is transferred into the gas.

Total work done = enclosed area / number of large squares approximately 40 (+/-5).
1 square = 5 J.
Work done = 200 J (+/-25 J).

ALTERNATIVE 1

«considering expansions from A» an adiabatic process will reduce/change temperature

and so curve AC must be the steeper

ALTERNATIVE 2

temperature drop occurs for BC

therefore CA must increase temperature «via adiabatic process».

Alternative 1:
Use the adiabatic formula $p_A V_A^{5/3}=p_C V_C^{5/3}$, so $V_C=\left(\frac{p_A}{p_C}\right)^{3/5}V_A$.
$V_C=\left(\frac{5.00\times10^5}{4.60\times10^3}\right)^{3/5}\times2.00\times10^{-3}=3.333\times10^{-2}\ \mathrm{m^3}$.

Alternative 2:
$V_C=V_B$ and $p_AV_A=p_BV_B$.
$V_C=\frac{5\times10^5\times2\times10^{-3}}{3\times10^4}$.

Alternative 3:
$V_C=V_B$ and $n=0.2\ \mathrm{mol}$.
$V_C=\frac{0.2\times8.31\times602}{3\times10^4}$.

For MP2, working or answer to at least 4 significant figures must be seen.

Increasing because thermal energy/heat is being provided to the gas « and temperature is constant, $\Delta S=\frac{\Delta Q}{T} \gg$

Alternative 1:
$Q=\Delta U=\frac{3}{2}V_C\Delta P$.
$Q=\frac{3}{2}\times3.33\times10^{-2}\times(3.00\times10^4-4.60\times10^3)=1268.7\approx1270\ \mathrm{J}$.

Alternative 2:
$Rn=\frac{5\times10^5\times2\times10^{-3}}{602}=1.66$ OR $T_C=\frac{4.6\times10^3\times3.33\times10^{-2}}{1.66}=92.2\ \mathrm{K}$.
$\Delta U=\frac{3}{2}\times1.661\times(602-92.21)=1270\ \mathrm{J}$.

Award [2] for BCA. Accept negative values.
Award MP1 if $T_C=92$ is taken from (e).

$$e_{\mathrm{C}}=1-\frac{92}{602}=0.847$$

this engine has $e<e_{\mathrm{c}}$ as it should

Marking guidance:

Award [0] if no calculation shown.