Calculate bulk modulous of water given that its volume changes from 100L to 100.5L at NTP.
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★Initial volume, V1 = 100.0l = 100.0 × 10 –3 m3
Final volume, V2 = 100.5 l = 100.5 ×10 –3 m3
Increase in volume, ΔV = V2 – V1 = 0.5 × 10–3m3
Increase in pressure, Δp = 100.0 atm = 100 × 1.013 × 105 Pa
Bulk modulus = Δp / (ΔV/V1) = Δp × V1 / ΔV
= 100 × 1.013 × 105 × 100 × 10-3 / (0.5 × 10-3)
= 2.026 × 109 Pa
Bulk modulus of air = 1 × 105 Pa
∴ Bulk modulus of water / Bulk modulus of air = 2.026 × 109 / (1 × 105) = 2.026 × 104
This ratio is very high because air is more compressible than water.★
______________________________★
★Hope this will help you»»««
Here is your answer»»
______________________________
★Initial volume, V1 = 100.0l = 100.0 × 10 –3 m3
Final volume, V2 = 100.5 l = 100.5 ×10 –3 m3
Increase in volume, ΔV = V2 – V1 = 0.5 × 10–3m3
Increase in pressure, Δp = 100.0 atm = 100 × 1.013 × 105 Pa
Bulk modulus = Δp / (ΔV/V1) = Δp × V1 / ΔV
= 100 × 1.013 × 105 × 100 × 10-3 / (0.5 × 10-3)
= 2.026 × 109 Pa
Bulk modulus of air = 1 × 105 Pa
∴ Bulk modulus of water / Bulk modulus of air = 2.026 × 109 / (1 × 105) = 2.026 × 104
This ratio is very high because air is more compressible than water.★
______________________________★
★Hope this will help you»»««
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