Question

A system is composed of 7.14 grams of Ne gas at 298 K and 1 atm. When 2025 joules of heat are added to the system at constant pressure, the resultant expansion causes the system to perform 810 joules of work. Calculate the following. (a) The initial state variables (P, V, T) (b) The final state variables. (c) The change in internal energy for the process. The molecular weight of Ne is 20 and you can assume Ne behaves as an ideal gas. (

Answer 1

Answer:

(a) 1 atm, 8.72 L and 298 K respectively.

(b) 1 atm, 16.72 L and 573.35 K respectively.

(c) $\Delta U=1215J$

Explanation:

Hello,

(a) In this case, considering that neon could be considered as an ideal gas, we can compute its volume as follows:

$PV=nRT\\\\V=\frac{nRT}{P}=\frac{7.14g*\frac{1mol}{20g}0.082\frac{atm*L}{mol*K}*298K}{1atm}\\ \\V=8.72L$

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:

$W=P(V_2-V_1)$

Thus, the final volume is:

$V_2=\frac{W}{P} +V_1=\frac{810Pa*m^3}{1atm*\frac{101325Pa}{1atm} } *\frac{1000L}{1m^3}+8.72L\\ \\V_2=16.72L$

And the final temperature is computed by considering the pressure-constant specific heat of neon (1.03 J/g*K) and the added heat:

$Q=mCp(T_2-T_1)\\\\T_2=\frac{Q}{mCp}+T_1 =\frac{2025J}{7.14g*1.03J/(g*K)}+298K\\ \\T_2=573.35K$

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:

$Q-W=\Delta U\\\\\Delta U=2025J-810J\\\\\Delta U=1215J$

Best regards.