Question 23M.2.SL.TZ1.7b

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 ✓