As a quantity of ideal gas expands isothermally from 0.0180 m3 to 0.0320 m3, it does 5.80×103 J of work. If th?
As a quantity of ideal gas expands isothermally from 0.0180 m3 to 0.0320 m3, it does 5.80×103 J of work. If the amount of gas is 3.79 mol, what is the temperature of the gas?
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Use the equation relating work with the isothermal process:
W = nRT ln (Vf/Vi)
W = 5.80*10^3 J
n - number of moles = 3.79 mol
R = gas constant, 8.31
Vf = 0.032 m3
Vi = 0.0180
Plug in the numbers and solve for T
5.80*10^3 =3.79 mol*(8.31)*T*In (0.0320/0.0180)
5.80*10^3 = 31.5 * T * In (1.78)
184 = T * In (1.78)
319 K = T
Simple ideal gas law...
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