Question

To 1 kg of argon (considered as an ideal gas), 2000 J of heat is added at constant P=1 atm. Calculate ΔU, ΔH, ΔT, and ΔV of the gas, and the work (Use CV =1.5R , CP = 2.5R)

Answer 1

ΔU = CVdT

ΔH= CPdT

Q= 2000 J

According to first law of thermodynamic

ΔU =Q+W

0=Q+W

Q=-W

-W= 2000 J

W= -2000 J

1.0 L atm = 101.3 J

2000 J * 1.0 L atm / 101.3 J= - 19.7 L atm

W= -pdV

Here p = 1.0 atm

-19.7 L atm = -1.0 atm dV

dV = 19.7 L

dH = dU+PdV

dU = 0,

dH = PdV = 19.7 L atm

molar mass of Ar = 39.948 g/ mol

number of moles = 1.0 kg or 1000 g/ 39.948 g/ mol = 25 mol

now for 25 mol

dH = 2* 19.7 L atm =493.1 4 L atm

R = 0.08206 L atm / K mol

CV =1.5R , CP = 2.5R

ΔU = CVdT

ΔH= CPdT

493.1 4 L atm = 2.5 * 0.08206 L atm / K

dT = 2403.8 K

ΔU = CVdT

=1.5 * 0.08206 L atm / K * 2403.8 K

= 295.88 L atm