In this problem, 0.99 mol of an ideal diatomic gas is heated at constant volume
ID: 1614963 • Letter: I
Question
In this problem, 0.99 mol of an ideal diatomic gas is heated at constant volume from 312 K to 564 K. (a) Find the increase in the internal energy of the gas, the work done by the gas, and the heat absorbed by the gas. internal energy work done heat absorbed (b) Find the same quantities if this gas is heated from 312 K to 564 K at constant pressure. Use the first law of thermodynamics and your results for Part (a) to calculate the work done by the gas. internal energy work done heat absorbed (c) Calculate the work done in Part (b). This time calculate it by integrating the equation dW = P dV.Explanation / Answer
The specific heat for ideal diatomic mass , Cv = 20.8 j/mol*k
The change in internal energy for this ideal di-atomic gas is the difference between the heat absorbed by the gas
and the Work it does.
The heat input into the system is:
Q = nCvT
= 0.99x[20.8]x[564-312]
=5189.184 Joules [Heat absorbed]
Work done
W = 0 [at constant volume]
internal energy :
E = Q - W
= 5189.184 -0
= 5189.184 joules
(b)
At constant pressure
:Cp = Cv+R = 20.8 +8.134
= 28.934 j/mol*K
Q = nCpT
= 0.99x28.934x[252]
= 7218.45 joules[ Heat]
W= nRT
= 0.99x8.314x252
= 2074.176 joules [ Work done]
E = Q - W
= 5144.27 joules[ Internal Energy]
(c)
dW = PdV
after integrating we get
W = P[V-U]
= nR[T2- T1]
= 0.99x8.314x252
= 2074.176 joules
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