Answer :
Answer:
The number of moles of the gas is: -27.14 mole
the charge in internal energy of the gas is -1.84 × 10⁴ J.
The work done by the gas is 4.42 × 10⁴ J
The heat liberated by the gas for the same temperature change while the volume was constant and is same as the change in internal energy.
As such ; Q = -1.84 × 10⁴ J.
Explanation:
The expression for the number of moles of a gas at constant pressure is as follows:
[tex]\mathbf{n = \frac{Q}{Cp \Delta T}}[/tex]
[tex]\mathbf{n = \frac{Q}{Cp (T_2-T_1)}}[/tex]
where ;
[tex]C_p[/tex] is the specific heat at constant pressure of Nitrogen gas which is = 29.07 J/mol/K
Since heat is liberated from the gas ; then:
[tex]n = \dfrac{-2.58*10^4 }{29.07(45-12.3)}[/tex]
n = -27.14 mole
The number of moles of the gas is: -27.14 mole
b) The expression to be used in order to determine the change internal energy is:
[tex]dU = nCv \Delta T[/tex]
where ;
n= 27.14 mole
Cv = specific heat at constant volume of Nitrogen gas = 20.76 J/mol/K
ΔT = (12.3-45)
So;
dU = (27.14)(20.76)(12.3-45)
dU = 563.426(-32.7)
dU = -18424.04328
dU = -1.84 × 10⁴ J
Thus; the charge in internal energy of the gas is -1.84 × 10⁴ J.
c) The workdone by the gas can be calculated as;
W = Q - ΔU
W = 2.58 × 10⁴ J - (-1.84 × 10⁴ J )
W = 2.58 × 10⁴ J + 1.84 × 10⁴ J
W = 4.42 × 10⁴ J
The work done by the gas is 4.42 × 10⁴ J
d) The expression to calculated the work done is given as:
W = pdV
since the volume is given as constant ; then dV = 0
so;
W = p(0)
W = 0
Replacing 0 for W in the equation W = Q - Δ U
0 = Q - ΔU
-Q = - ΔU
Q = ΔU
Thus , the heat liberated by the gas for the same temperature change while the volume was constant and is same as the change in internal energy.
As such ; Q = -1.84 × 10⁴ J.