Compression of Gases
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How much work must be done to 30 grams of CO at standard temperature and pressure to compress it to 1/5 of its initial volume if the process is a) isothermal b) adiabatic?
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Solution Summary
This solution provides step-by-step equations for determining work needed to compression a gas. The standard temperature and pressures compressed for initial volume processes are analyzed. Isothermal and adiabatic processes are examined.
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CO is a diatomic gas.
gamma = Cp/Cv
For a diatomic gas, gamma = 1.4
The mass of CO = 30 grams
So the number of moles of CO, n = wt/atomic mass = 30/28 = 1.07
a) isothermal
Work done = 2.3026 x n x R x T x Log (Vf/Vin)
where, Vf = final volume = (1/5)Vin
Vin = ...
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