Hence,
Q = H = n(5/2)RT
=>
T = Q/( n(5/2)R )
= 43.08J / (3.00410mol (5/2) 8.314472J/molK )
= 69K = 69 C
(a)
Since no work is done in a constant volume process, the change of internal energy equals the heat transferred to the gas.
U = Q
Assuming ideal gas behavior, the internal energy of the argon sample ideal gas is given by:
U = nCvT
thus,
U = nCvT
The molar heat capacity at constant volume of a monatomic ideal gas like argon is:
Cv = (3/2)R
The number of moles in the sample is
n = m/M = 1.2g / 39.948g/mol = 3.00410mol
Hence,
Q = U = n(3/2)RT
= 3.00410mol (3/2) 8.314472J/molK 115K
= 43.08J
(b)
The heat transferred in a constant pressure process equals the change of enthalpy.
H = Q
Internal energy of an ideal gas is given by:
H = nCpT
<=>
H = nCpT
The molar heat capacity at constant pressure of a monatomic ideal gas is:
Cp = (5/2)R
Hence,
Q = H = n(5/2)RT
=>
T = Q/( n(5/2)R )
= 43.08J / (3.00410mol (5/2) 8.314472J/molK )
= 69K = 69 C
This post is last updated on hrtanswers.com at Date : 1st of September – 2022