Question

A petroleum engineer has a tank with 2.00mol of N2 gas at 1.00 atm and 16.0?C . She first heats the gas at constant volume, adding 1.52

Answer 1

n=2mol

p= 1atm= 101325 Pa

T= 16+273= 289 K

so V_initial= nRT/P= 0.0474 m³

the process is isochoric so V₂=V_initial= 0.0474 m³

now heat added= 1.52*10⁴ J

using first law of thermodynamics

U= q-w

so w= 0 because the process is isochoric

U= q= nCv T

and Cv= (5/2)R

therefore

2 *(5/2)R*(T2-T1)=1.52*10⁴ J= 15.2 KJ

putting R=8.314 and T1=289 K

we get T2= 654.6 K

P2= nRT2/V2= 2.296 *10⁵ Pa

now part 2 begins

its isobaric

so P3= P2= 2.296 *10⁵ Pa

and V3= 2*V2=2*0.0474 m³ = 0.0948 m³

work done by gas= P(V3-V2)= 10.883 kJ= 10883J

T3= P3 V3/nR= 1636.196 K

so temprature at the end= 1309.0016 K

amount of heat added

by first law,

U+w=q

U= n(5/2)R(T3-T2)= 27.203 KJ

and work done= 10.883 kJ

q=  10.883 kJ+ 27.203 KJ= 38.086478 kJ

Heat added to the gas= 38.086478 kJ

net change in the internal enegy= change in internal energy in step1+ change in internal energy in step2

= 15.2 KJ+27.203 KJ=42.403 KJ

net change in the internal enegy=42.403 KJ