Answer:
(a) 1 atm, 8.72 L and 298 K respectively.
(b) 1 atm, 16.72 L and 573.35 K respectively.
(c) [tex]\Delta U=1215J[/tex]
Explanation:
Hello,
(a) In this case, considering that neon could be considered as an ideal gas, we can compute its volume as follows:
[tex]PV=nRT\\\\V=\frac{nRT}{P}=\frac{7.14g*\frac{1mol}{20g}0.082\frac{atm*L}{mol*K}*298K}{1atm}\\ \\V=8.72L[/tex]
Thus, initial conditions of pressure volume and temperature are 1 atm, 8.72 L and 298 K respectively.
(b) Since the process was carried out at constant pressure, the work is defined as:
[tex]W=P(V_2-V_1)[/tex]
Thus, the final volume is:
[tex]V_2=\frac{W}{P} +V_1=\frac{810Pa*m^3}{1atm*\frac{101325Pa}{1atm} } *\frac{1000L}{1m^3}+8.72L\\ \\V_2=16.72L[/tex]
And the final temperature is computed by considering the pressure-constant specific heat of neon (1.03 J/g*K) and the added heat:
[tex]Q=mCp(T_2-T_1)\\\\T_2=\frac{Q}{mCp}+T_1 =\frac{2025J}{7.14g*1.03J/(g*K)}+298K\\ \\T_2=573.35K[/tex]
Therefore, final volume is 16.72 L, final pressure is also 1 atm and final temperature is 573.35 K for this expansion process.
(c) Finally, the change in the internal energy is computed via the first law of thermodynamics:
[tex]Q-W=\Delta U\\\\\Delta U=2025J-810J\\\\\Delta U=1215J[/tex]
Best regards.
