In a constant-volume process, 208 J of energy is transferred by heat to 0.96 mol of an ideal monatomic gas ini?
In a constant-volume process, 208 J of energy is transferred by heat to 0.96 mol of an ideal monatomic gas initially at 293 K.
(a) Find the increase in internal energy of the gas.
(b) Find the work done on it.
(c) Find its final temperature.
Please show me the steps(formulas) as well. Thanks
- schmisoLv 71 decade agoFavorite Answer
The change of internal energy equals the energy transferred to the gas by work and heat.
∆U = W + Q
In a constant volume process no work is done. Therefore:
∆U = Q = 208J
As point out in (a)
W = 0
Internal energy of an ideal gas is given by:
U = n∙Cv∙T
Molar heat capacity at constant volume for an monatomic ideal gas is:
Cv = (5/3)∙R
∆U = n∙(5/3)∙R∙∆T = n∙(5/3)∙R∙(T₂ - T₁)
T₂ = T₁ + 3∙∆U/(5∙n∙R)
= 293K + 3∙208J/(5 ∙ 0.96 ∙ 8.314472J/molK)
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