2 mole of an ideal gas at 2atm and 300K is compressed at isothermal to one-half of its volume by a constant external pressure of 4 atm.calculate q, w,and ∆E?
(R=0.082 atm lit/mol cal)
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n= 2
V1 = V
V2 = V/2
T = 300
For an isothermal process
W = nRT ln[ V2/V1]
= 2 * 0.082 * 300 ln [1/2]
= 33.948 J
in an isothermal process since temprature is constant
thus
U = 0
And thus
Q= - W
= - 33.948
V1 = V
V2 = V/2
T = 300
For an isothermal process
W = nRT ln[ V2/V1]
= 2 * 0.082 * 300 ln [1/2]
= 33.948 J
in an isothermal process since temprature is constant
thus
U = 0
And thus
Q= - W
= - 33.948
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