Neon is a monatomic gas.

The ratio of the specific heats is γ = 5/3 for monatomic ideal gas and cV = 3R/2

We will start by calculating TA using the ideal gas law. The ideal gas law states pV = nRT

where

p is the pressure,
V is the volume,
n is the number of moles of gas,
R is the gas constant,
and T is the temperature in Kelvin.

From the graph VA = 400 × 10-6 m3 and pA = 2.50 × 105 Pa

From the graph, VB = 400 × 10-6 m3 and pB = 2.50 × 105 Pa

For an adiabatic process, pVγ = constant.

Therefore,

Substituting with the given and solving for VC

To calculate TC we apply the ideal gas law

Now we can calculate the work done, the heat transferred, and the variation of the thermal energy during each transformation

The work done by the system (Ws) is the area enclosed under the p-V curve on the diagram.
The variation of the internal energy ΔE = nCvΔT
The heat Q is calculated using the first law of thermodynamic ΔE = Q-WS
Q is the heat added to the system Q = ΔE+WS

Transformation AB

Ws,AB = 0 J
ΔEAB = nCv(TB - TA) = nCvΔT = (0.0100 mol)(3/2)(8.31 J/mol•K)(1.20 ×103 K - 7.22 ×102 K) = 59.6 J
QAB = 60 J+0 J = 59.6 J

Transformation BC (Adiabatic)

QBC=0 J (Adiabatic)
ΔEBC=nCv(TC - TB) = nCvΔT = (0.010 mol)(3/2)(8.31 J/mol•K)(9.80 ×102 K - 1.20 × 103 K) = - 27.4 J

Transformation CA

W_s,CA= p_C(V_C-V_A)=1.50 × 10⁵ Pa × (543 ×10^-6 m³ - 400 ×10^-6 m³) = 21.5 J
Δ_ECA=nC_vΔT=(0.01 mol)(3/2)(8.31 J/mol•K)(9.80 ×10² K - 7.22 ×10² K) = 32.2 J
Q_CA= 21 J + 32J = 53.7 J

During the cycle:

the net work done by the gas is Ws,net=Ws,AB+Ws,BC+Ws,CA,
the added heat energy added to the gas is Q=QAB+QBC+QCA
the variation in the thermal energy during the cycle is ΔE = ΔEAB+ΔEBC+ΔECA

The net work done by the gas, the added heat energy added to the gas, and the variation in the thermal energy during the cycle are Ws,net = 5.9 J, Q = 5.9 J, and ΔE = 0 J, respectively.

Cyclic Thermodynamic Processes Practice Problems