A block of metal whose volume expansivity is 5.0 × 10−5 °C−1 and isothermal compressibility is 1.2 × 10−6 atm−1 has volume 5 litre at 1 atm and 20 °C. On applying ... Calculate the pressure applied
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Due to increase in pressure the volume decreases. Due to increased temperature volume increases. So the net increase is 0.5 cc given.
Expected expansion in volume due to ΔT = 12° C is
= 5000 cc * 5.0 * 10⁻⁵ /⁰C * 12°C = 3 cc
Actual increase = 0.5 cc
So compression in volume due to ΔP = 2.5 cc
Compressibility = - 1/V * ΔV / ΔP = β_T = 1.2 * 10⁻⁶ /atm
ΔP = 2.5 / [ 5000 * 1.2 * 10⁻⁶) = 416.67 atm
Pressure applied = P + ΔP = 417.67 atm
Expected expansion in volume due to ΔT = 12° C is
= 5000 cc * 5.0 * 10⁻⁵ /⁰C * 12°C = 3 cc
Actual increase = 0.5 cc
So compression in volume due to ΔP = 2.5 cc
Compressibility = - 1/V * ΔV / ΔP = β_T = 1.2 * 10⁻⁶ /atm
ΔP = 2.5 / [ 5000 * 1.2 * 10⁻⁶) = 416.67 atm
Pressure applied = P + ΔP = 417.67 atm
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