Answer :
Answer:
63.5 °C
Explanation:
The expression for the calculation of work done is shown below as:
[tex]w=P\times \Delta V[/tex]
Where, P is the pressure
[tex]\Delta V[/tex] is the change in volume
Also,
Considering the ideal gas equation as:-
[tex]PV=nRT[/tex]
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 8.314 J/ K mol
So,
[tex]V=\frac{nRT}{P}[/tex]
Also, for change in volume at constant pressure, the above equation can be written as;-
[tex]\Delta V=\frac{nR\times \Delta T}{P}[/tex]
So, putting in the expression of the work done, we get that:-
[tex]w=P\times \frac{nR\times \Delta T}{P}=nR\times \Delta T[/tex]
Given, initial temperature = 28.0 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (28.0 + 273.15) K = 301.15 K
W=1770 J
n = 6 moles
So,
[tex]1770\ J=6 moles\times 8.314\ J/ Kmol \times (T_2-301.15\ K)[/tex]
Thus,
[tex]T_2=301.15\ K+\frac{1770}{6\times 8.314}\ K[/tex]
[tex]T_2=336.63\ K[/tex]
The temperature in Celsius = 336.63-273.15 °C = 63.5 °C
The final temperature is:- 63.5 °C