Question

Consider a process that occurs at constant volume.

The initial volume of gas is   2.30 L  , the initial temperature of the gas is   26.0 °C  , and the system is in equilibrium with an external pressure of 1.2 bar (given by the sum of a 1 bar atmospheric pressure and a 0.2 bar pressure due to a brick that rests on top of the piston). The gas is heated slowly until the temperature reaches  51.2 °C  . Assume the gas behaves ideally, and its heat capacity at constant volume is:  CV,m=14.2J/(K.mol)

Part A

What is the value of w? (J)

Part B

What is the value of q? (J)

Part C

Calculate  ΔU

Part D

Calculate  Δ H . Remember the definition of enthalpy: H=U+pV, which applied to a process becomes  ΔH=ΔU+Δ(pV).

Answer 1

part A)
since process occures at constant volume
so, W = 0

part C)
delta U = n*Cv*delta T
n = number of mole
= (P*V)/(R*T)
T1 = 26oC
= (273+26) k
= 299 k
n = (1.2*2.3)/(0.082*299)
= 0.113 mole
T2 = 512 oC
= (273+512) k
= 785 k

delta U = 0.113*14.2*(785-299)
= 779.8 J

part B)
since, W = 0
so, delta U = q
q = 779.8 J

part D)
delta H = delta U + delta(PV)
delta H = delta U + V*deltaP + P*delta V

at constant volume
P1/T1 = P2/T2
1.2/299 = P2/785
P2 = 3.15 bar
= (3.15*10^5) Pa
delta V = 0
delta H = 779.8 + 2.3*(3.15-1.2)*10^5 + 0
delta H = 779.8 +448500
delta H = 449279.8 J