In first chamber, volume, pressure and temperature of the gas be \(V_{1}, P_{1} \ and \ T_{1}\)
In second chamber, volume, pressure and temperature of the gas be \(V_{2}, P_{2} \ and \ T_{2}\)
By conservation of energy, we get
\(\left(\frac{P_{1} V_{1}}{R T_{1}}+\frac{P_{2} V_{2}}{R T_{2}}\right) \mathrm{T} =\frac{P_{1} V_{1}}{R}+\frac{P_{2} V_{2}}{R}  \)
\(\text {Internal energy } =\frac{3}{2}\left[n_{1} R T_{1}\right]+\frac{3}{2}\left[n_{2} R T_{2}\right]  \)
\(\therefore \frac{n_{1} T_{1}+n_{2} T_{2}}{n_{1}+n_{2}} =\left(n_{1}+n_{2}\right) C_{\mathrm{v}} T  \)
\(=n_{1} C_{v} T+n_{2} C_{v} T_{2} \)
\(\therefore\left(\frac{P_{1} V_{1} T_{2}+P_{2} V_{2} T_{1}}{T_{1} T_{2}}\right) \mathrm{T} =\mathrm{P}_{1} \mathrm{~V}_{1}+\mathrm{P}_{2} \mathrm{~V}_{2}  \)
\(\therefore \mathrm{T} =\frac{T_{1} T_{2}\left(P_{1} V_{1}+P_{2} V_{2}\right)}{P_{1} V_{1} T_{2}+P_{2} V_{2} T_{1}}\)

Normal temperature of human body
\(\mathrm{T}_{1}=98.6^{\circ} \mathrm{F}\)
Final temperature of human body
\(T_{2}=104^{\circ} \mathrm{F}\)
Peak wavelength
\(\lambda_{m}=\frac{b}{T}  \)
\(\mathrm{~T}_{1}=98.6+273=371.6 \mathrm{~K}\)
Peak wavelength at \(98.6^{\circ} \mathrm{F}\)
\(\lambda_{m} =\frac{2.898 \times 10^{-8}}{371.6}  \)
\(=0.007798 \times 10^{-8}  \)
\(\lambda_{\max } =7798 \mathrm{~nm}  \)
\(\mathrm{~T}_{2} =104+273 \)
= 377 K
Peak wavelength at \(104^{\circ} \mathrm{F}\)
\(\lambda_{m} =\frac{2.898 \times 10^{-8}}{377}  \)
\(=0.007687 \times 10^{-8} \)
= 7687 nm

P₁ = 1 atmospheric Pressure
V₁ = V
V₂ = V/8
T₁ = 32 + 273
= 305
T₂ = ? 
\(\gamma =1.4  \)
\(T_{2} V_{2}^{\gamma-1} =T_{1} V_{1}^{\gamma-1}  \)
\(\frac{T_{1}}{T_{2}} =\left(\frac{V_{2}}{V_{1}}\right)^{\gamma-1}  \)
\(\frac{T_{1}}{T_{2}} =\left(\frac{V / 8}{V}\right)^{1.4-1}  \)
\(=\left(\frac{1}{8}\right)^{0.4} \)
\(\frac{305}{T_{2}} =\frac{1}{8^{0.4}}\)
\(\therefore T_{2} =305 \times 8^{0.4}  \)
\(=400^{\circ} \mathrm{C}\)
Temperature of the compressed air
\(T \cong 400^{\circ} \mathrm{C}\)
 

(a) Work done by the gas along AB
W = P\(\triangle \)V
∴ W = 600\(\times\)3 = 1800 J = 1.8 kJ
(b) Work is done on the gas along BC 
W = -P\(\triangle \)V
= -400\(\times\)(6 - 3)
= - 400 x 3 = -1200 J
=-1.2 kJ
(c) Net work done in the process
= Area under the curve AB
= Area of rectangle + Area of triangle
Area of triangle =\(\cfrac { 1 }{ 2 } \)\(\times\)b\(\times\) h
\(=\cfrac { 1 }{ 2 } \times \triangle V\times \triangle P\)
=\(\cfrac { 1 }{ 2 } \)\(\times\)3\(\times\)200 = 300J
Area of rectangle = I\(\times\) b
=400\(\times\)3 = 1200 J
∴ Net work done= 1200 + 300
W = 1500 J = 1.5 J
∴ W = 1.5 J

a. Isochoric \(\Rightarrow \)V = V₀ = constant

b. Isothermal \(\Rightarrow \) T = T_o = constant
PV = nRT_o                         P_V = nRT 

P(T) = multivalued
c. Isobaric \(\Rightarrow \)P = P_o = constant
P_oV = nRT