Question

The PV diagram shows the compression of 60.1 moles of an

ideal monoatomic gas from state A to state B. Calculate Q,

the heat added to the gas in the process A to B. Data: PA=

1.84E+5 N/m2 VA= 1.97E+0 m3 PB= 1.15E+5 N/m2 VB=

9.30E-1 m3

Answer 1

Q = deltaU + W

What is deltaU?

Well, Ua and Ub...

deltaU = Ub - Ua

To find the value of U at both states, use the calorically perfect assumption:

Ua = n*cv*Ta

Ub = n*cv*Tb

What is molar cv in terms of more easy to remember values?

cv = R/(k - 1)

where k is the adiabatic index. Later we will assign k=5/3 because it is a monatomic gas.

Thus:

deltaU = n*R*(Tb - Ta)/(k - 1)

Recognize ideal gas law, such that:

Ta = Pa*Va/(n*R)

Tb = Pb*Vb/(n*R)

Thus:

deltaU = (Pb*Vb - Pa*Va)/(k - 1)

So we can develop our expression for Q:

Q = (Pb*Vb - Pa*Va)/(k - 1) + 1/2*(Pb + Pa)*(Vb - Va)

If you care, our resulting expressions for work and internal energy change:

W = 1/2*(Pb + Pa)*(Vb - Va)

deltaU = (Pb*Vb - Pa*Va)/(k - 1)

Data:

Pa:= 183 kPa; Pb:=115 kPa; Va:=1.97 m^3; Vb:=0.930 m^3;

W= -154.96 , deltaU= -380.34

Result:Q = -535.3 kiloJoules, negative sign indicates gas rejects heat to the surroundings