First slide

Thermodynamic processes

Question

Logarithms of readings of pressure and volume for an ideal gas were plotted on a graph as shown in the given figure. By measuring the gradient, it can be shown that the gas may be

Moderate

A

monatomic and undergoing an adiabatic change
B

monatomic and undergoing an isothermal change
C

diatomic and undergoing an adiabatic change
D

triatomic and undergoing an isothermal change

Solution

$\log P = m \log V + C_{1}$ , where C₁ is positive, m is slope
$m = \frac{2 .38 - 2 .10}{1 .1 - 1 .3} = - 1 .4$
$\log P = - 1 .4 \log V + C_{1}$
$\log {PV}^{1 .4} = C_{1}$
${PV}^{1 .4} = k$
Thus, it represents an ideal diatomic gas undergoing adiabatic change.

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