Question 23M.3.SL.TZ1.7

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 of adiabatic formula «$p_{\mathrm{A}}{V_{\mathrm{A}}}^{\frac{5}{3}}=p_{\mathrm{C}}{V_{\mathrm{C}}}^{\frac{5}{3}} \Rightarrow$» $V_{\mathrm{C}}={\left(\frac{p_{\mathrm{A}}}{p_{\mathrm{C}}}\right)}^{\frac{3}{5}}{V}_{\mathrm{A}}$ ✓

$V_{\mathrm{C}}={\left(\frac{5.00\times {10}^5}{4.60\times {10}^3}\right)}^{\frac{3}{5}}$ × 2.00 × 10⁻³ «= 3.333 × 10⁻² m³» ✓

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

ALTERNATIVE 2

V_C=V_B AND p_A V_A = p_B V_B ✓

$V_{\mathrm{C}}=\frac{5\times {10}^5\times 2\times {10}^{-3}}{3\times {10}^4}$ ✓

ALTERNATIVE 3

V_C=V_B AND n = 0.2 mol ✓

V_C = (0.2 × 8.31 × 602) / 4 × 10⁴ ✓

Increasing ✓

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

ALTERNATIVE 1

$Q=\mathrm{\Delta }U=\frac{3}{2}{V}_{\text{C}}\mathrm{\Delta }P$ ✓

$Q=$«$\frac{3}{2}$ × 3.33 × 10⁻² × (3.00 × 10⁴ − 4.60 × 10³)» = 1268.7 ≈ 1270 «J» ✓

Award [2] for BCA.

Accept negative values.

ALTERNATIVE 2

$Rn=\frac{5\times {10}^5\times 2\times {10}^{-3}}{602}=1.66$  OR  T_c = 4.6 × 10³ × 3.33 × 10⁻² × 1.66 = 92.2  ✓

$\mathrm{\Delta }U=\frac{3}{2}\times 1.661\times (602-92.21)=1270$ «J» ✓

Award MP1 if T_c = 92 taken from (e)

e_c = 1 − $\frac{92}{602}$ = 0.847 ✓

this engine has e < e_c as it should ✓

Award [0] if no calculation shown.