Problem 2.2 The temperature of 1.30 mol of an ideal gas increases from 10.4 celc

Question

Problem 2.2 The temperature of 1.30 mol of an ideal gas increases from 10.4 celciusto 48 .8 celcius as the gas is compressed adiabatically. Part A Calculate q for this process, assuming that .C=3R/2 Express your answer to one significant figure and include the appropriate units. = Part B Calculatew for this process, assuming that .C=3R/2 Express your answer to three significant figures and include the appropriate units. = Part C Calculate delta U for this process, assuming that .C=3R/2 Express your answer to three significant figures and include the appropriate units. = Part D Calculate delta H for this process, assuming that .C=3R/2 Express your answer to three significant figures and include the appropriate units. =

Explanation / Answer

You use equations for an ideal gas: pV = NkT = nRT U = CvNkT = Cv nRT where: Cv is a constant dependent on temperature (e.g. equal to 3/2 for a monatomic gas for moderate temperatures) U is the internal energy p is the pressure V is the volume n is the amount of gas (moles) R is the gas constant, 8.314 J·K^-1mol^-1 T is the absolute temperature N is the number of particles k is the Boltzmann constant, 1.381×10^-23 J·K^-1 In your first problem, the energy difference between heat absorbed by the gas and the work performed is the energy lost to heat. Thus, ?E = Cv nR?T where ?T is the temperature change ?T = ?E / (Cv nR) = (750-625) / (3/2)(1.3)(8.314) = 7.71° Kelvin

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