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smilebug73
10.12.2019 •
Physics
Suppose 8.41 moles of a monatomic ideal gas expand adiabatically, and its temperature decreases from 395 to 279 k. determine (a) the work done (including the algebraic sign) by the gas, and (b) the change in its internal energy.
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Ответ:
a) W=12166.20876 J
b) U= -12166.20876 J
Explanation:
No. of moles, n = 8.41
Change of temperature, ΔT = T1 - T2
= 395 - 279
= 116 K
For monatomic gas, γ = 5/3
γ -1 = 2 /3
Solution:
(a)
Work done,![W= \frac{nR}{\gamma-1}(T_1-T_2)](/tpl/images/0412/2626/aa9a3.png)
plugging values we get
Ans: 12166.20876 J
Work done, W = + 12166.20876 J
(b)
From first law of thermodynamics, dQ = U + W
but, dQ = 0 ( adiabatic process)
Hence, U = - W
= - 12166.20876 J
Ans:
Change in internal energy, U = - 12166.20876 J
Ответ:
a lot of MASS
Explanation: