- #1

roam

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## Homework Statement

http://img29.imageshack.us/img29/4386/prob1p.jpg

## The Attempt at a Solution

To answer the first question as to why the air needs to be refrigerated, I think it's because refrigerating the gas cause it to do work, so its enthalpy will decrease resulting in a decrease in temperature. Is the correct?

This is my attempt at the calculation part: Since this is an Adiabatic process we consider the case, Q = 0 (when the heating or cooling is zero). So du = c

_{v}dT=-pdα. To use this equation we need to know the density since α=1/ρ.

I first found the initial volume using the equation PV/T = nR, and since nR is constant (no gas escapes):

[itex]\frac{P_iV_i}{T_i} = \frac{P_fV_f}{T_f}[/itex]

[itex]\frac{(300 \times 10^2)V_i}{-40} = \frac{(800 \times 10^2) \times 0.0005 \ m^2}{20}[/itex]

[itex]\therefore V_i = - 0.00266 m^3[/itex]

I don't understand the -ve in volume but:

[itex]\rho = \frac{PV}{RT} = \frac{(800 \ hPa -300 \ hPa) \times (0.0005+0.00266)}{287 \times (-40+20)} = -0.0275 \ Kg/m^3[/itex]

[itex]du = -(800 \times 10^2) \frac{1}{-0.0275} = 2906329.114 \ J[/itex]

I don't think my calculation is correct because it didn't take into account the specific heat of air that they provided. If I used du = c

_{v}dT = 1005 x (20+40) = 60300, which is a completely different answer. Any help or hints are greatly appreciated.

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