You compress a gas in an insulated cylinder— no heat flows into or out of the gas. The gas pressure is fairly low, so treating the gas as ideal is a good approximation. When you measure the pressure as a function of the volume of the gas, you obtain these results:

(a) Graph log 1p2 versus log 1V2, with p in Pa and V in m3 . Explain why the data points fall close to a straight line. (b) Use your graph to calculate g for the gas. Is the gas monatomic, diatomic, or polyatomic? (c) When p = 0.101 atm and V = 2.50 L, the temperature is 22.0o C. Apply the ideal-gas equation and calculate the temperature for each of the other pairs of p and V values. In this compression, does the temperature of the gas increase, decrease, or stay constant?

It is a thermodynamic process where no heat is exchanged between system and surroundings.

Q=0

a)

We have,

In adiabatic process,

$$\begin{gathered} {\mathrm{PV}}^{\mathrm{\gamma }}=\mathrm{constant} \\ {\mathrm{PV}}^{\mathrm{\gamma }}=\mathrm{A} \\ \mathrm{InP}+\mathrm{\gamma InV}=\mathrm{InA} \\ \mathrm{InP}=\mathrm{InA}‐\mathrm{\gamma InV} \end{gathered}$$ take logarithm,

Thus, as above equation is$\mathrm{InP}=\mathrm{InA}‐\mathrm{\gamma InV}$ comes with negative slope so the data points fall close to a straight line.

As,

$\mathrm{InP}=\mathrm{InA}‐\mathrm{\gamma InV}$

The slope is negative because (specific heat ratio) is negative.

$$\begin{gathered} m=‐\gamma \\ \gamma =‐m \\ \gamma =1.39 \\ \gamma ~1.4 \end{gathered}$$

Thus, the gas is monoatomic.

Use ideal gas law,

P=0.0101atm

V=2.5L

T=295.15K=22

Put all the values,

$$\begin{gathered} \mathrm{PV}=\mathrm{nRT} \\ \mathrm{nR}=\frac{\mathrm{PV}}{\mathrm{T}} \\ \mathrm{nR}=\frac{\left(0.101\mathrm{atm}\right)\left(2.5\mathrm{L}\right)}{295.15\mathrm{K}} \\ \mathrm{nR}=8.5550\times {10}^{-4}\mathrm{atm}\times \mathrm{L}/\mathrm{K} \end{gathered}$$

ii)

$$\begin{gathered} \mathrm{P}=0.139\mathrm{atm} \\ \mathrm{V}=2.02\mathrm{L} \\ \mathrm{PV}=\mathrm{nRT} \\ \mathrm{T}=\frac{\mathrm{PV}}{\mathrm{nR}} \\ \mathrm{T}=\frac{\left(0.139\mathrm{atm}\right)\left(2.02\mathrm{L}\right)}{8.5550\times {10}^{-4}\mathrm{atm}\times \mathrm{L}/\mathrm{K}} \\ \mathrm{T}=328.21\mathrm{K} \\ \mathrm{T}={55.056}^{\circ }\mathrm{C} \\ {\mathrm{T}}_2={55.056}^{\circ }\mathrm{C} \end{gathered}$$

iii)

$$\begin{gathered} \mathrm{PV}=nRT \\ T=\frac{\mathrm{PV}}{\mathrm{nR}} \\ \mathrm{T}=\frac{\left(0.202\mathrm{atm}\right)\left(1.48\mathrm{L}\right)}{8.5550\times {10}^{-4}\mathrm{atm}\times \mathrm{L}/\mathrm{K}} \\ \mathrm{T}=349.46\mathrm{K} \\ {\mathrm{T}}_3={76.306}^{\circ }\mathrm{C} \end{gathered}$$

iv)

P=0.361atm

V=1.01L

PV=net

T=$\frac{PV}{nR}$

Put all the values,

T=426.20K

T₄=153.05

v)

P=0.952atm

V=0.50L

PV=nRT
T= $\frac{PV}{nR}$

Put all the values,

T=556.40K

T₅=283.25

Thus,

T(22.0^oC) < T₂(55.056^oC) < T₃(76.306^oC) < T₄(153.05^oC) < T₅(283.25^oC) temperature of gas increases as pressure increases.