Question

3. An ideal monatomic gas expands isothermally from .500 m3 to 1.25 m3 at a constant temperature of 675 K. If the initial pressure is 1.00 ∙ 105 Pa, find (a) the work done by the gas, (b) the thermal energy transfer Q, and (c) the change in the internal energy.

Answer #1

Initial volume V = 0.5 m ^{3}

Final volume V ' = 1.25 m ^{3}

Temperature T = 675 K

Initila pressure P = 1x10 ^{5} Pa

From the relation PV = nRT

Number of moles n = PV /RT

Where R = Gas constant = 8.314 J / mol K

Substitute values you get , n = (1x10
^{5})(0.5)/(8.314)(675)

= 8.909 mol

(a). Work done by the gas W = nRT ln(V ' / V )

W = (8.909)(8.314)(675) ln( 1.25/0.5)

= 50000 ln(2.5)

= 50000(0.9162)

= 45814.5 J

(b).In isothermal process ,the thermal energy transfer Q = W

So, Q = 45814.5 J

(c) In isothermal process ,the change in the internal energyU = 0

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