Question

Consider a process that occurs at constant volume.

The initial volume of gas is   1.50 L , the initial temperature of the gas is   30.0 °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  55.2 °C . Assume the gas behaves ideally, and its heat capacity at constant volume is:  CV,m=14.2J/(K.mol)

What is the value of w?  (think about the sign first)

w =

Part B

What is the value of q? (think about the sign first)

q =

Part C

Calculate  ΔU

ΔU =

Part D

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

ΔH =

Answer 1

Sol.

Part A :  

As volume is constant , so , change in volume ( deltaV ) is 0 So ,   work done = w = - P × deltaV = 0   

Part B :

As Initial Volume = V1 = 1.50 L

Initial Pressure = P1 = 1.2 bar = 1.2 × 0.987 atm

= 1.1844 atm

Initial Temperature = T1 = 30°C

= 30 + 273.15 K = 303.15 K   

Gas constant = R = 0.0821 L atm / K mol

So , Moles of gas = n = P1V1 / RT1  

= 1.1844 atm × 1.50 L / ( 0.0821 L atm / K mol × 303.15 K )

= 0.0714 mol

Final Temperature = T2 = 55.2°C

= 55.2 + 273.15 K = 328.35 K   

Heat Capacity at constant volume = Cv,m = 14.2 J / K mol

So , at constant volume ,

Change in internal energy

= deltaU = nCv,m( T2 - T1)  

= 0.0714 mol × 14.2 J / K mol × ( 328.35 K - 303.15 K )

= 25.5498 J  

According to Ist law of thermodynamics ,

Amount of heat changed = q = deltaU + w

As w is 0  

So , q = deltaU =   25.5498 J   

Part C :

deltaU =   25.5498 J   

Part D :

At constant Volume ,

P1 / T1 = P2 / T2  

where P1 and P2 are initial and final pressures respectively and T1 and T2 are initial and final temperatures respectively

As P1 = 1.1844 atm , T1 = 303.15 K , T2 = 328.35 K

So , P2 = (P1/T1) × T2

= ( 1.1844 atm / 303.15 K ) × 328.35 K   

= 1.2828 atm

Now , Change in enthalpy = deltaH

= deltaU + delta(PV)

= deltaU + deltaPV + PdeltaV

As volume is constant , so , deltaV = 0

and , deltaH = deltaU + deltaPV

Now , deltaPV = (P2-P1) × V1

= ( 1.2828 atm - 1.1844 atm ) × 1.50 L

= 0.1476 L atm  

= 0.1476 × 101.325 J

= 14.95557   J

Therefore ,

deltaH = 25.5498 J + 14.95557 J  

=   40.5054 J