Question 1 Report
d1 = 0.600g/ litre
θ2 = 350°C, T2 = 2 + 3 + θ
= 273 + 350 = 623k, d2 = ?
P2 = 7atm
T a P a 1/d = TPK/d (where k is a constant)
Tp1 ÷ d1 = T1P2 ÷ d2, [(308 × 5) ÷ 0.600]
= (623 × 7) ÷ d2
d2 = (623 × 7 ×0.600) ÷ (308 × 5)
= 2616.6 ÷ 1540
= 1.69g / litresd2 = 1.69g / litres
∴ d2 = 1.70g / litres (Approx.)
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