InterviewSolution
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InterviewSolution
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A cylinder contains 3 moles of oxygen at a temperature of `27^(@)C`. The cylinder is provided with a frictionless piston which maintains a constant pressure of 1 atm on the gas. The gas is heated until its temperature rises to `127^(@)C`. a. How much work is done by the gas in the process ? b. What is the change in the internal energy of the gas ? c. How much heat was supplied to the gas ? For oxygen `C_(P) = 7.03 cal mol^(-1) .^(@)C^(-1)`. |
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Answer» For oxygen `C_(P) = 7.03 xx cal//mol .^(@)C`. `Delta Q = nC_(P) Delta T = 3 xx 7.03 xx (127 - 27)` `= 2109 cal (n = 3)` Since `gamma = C_(P)//C_(v)` and `gamma = 1.4` for oxygen as it is diatomic, `Delta U = nC_(v) Delta T = 3 xx (7.03)/(1.4) xx 100 = (2109)/(1.4) cal = 1506 cal` By the first law of thermodynamics, `Delta Q = Delta U + Delta W` `:. Delta W = Delta Q - Delta U = 2109 = 1506 = 603 cal` `= 603 xx 4.184 = 2523 J` a. work done by the gas `= 2523 J` b. Change in internal energy `= 1506 cal` c. Heat added to the gas ` = 2109 cal` |
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