Question

A monoatomic ideal gas is taken in a cylinder. It undergoes through the process $A B C$ as shown in figure. If the temperature at the point $A$ is $300 K$ Find

(i) the temperature at points $B$ and $C$ .

(Use gas constant $R = \frac{25}{3} J / mol - K$ )

Accepted Answer

The number of moles of gas, as $n = \frac{P_{A} V_{A}}{R T_{A}}$ $. . . (i)$

We are given: Initial state pressure, volume and temperature of gas as $P_{A} = 1 \times 10^{5} Pa, V_{A} = 100 \times 10^{- 6} m^{3}$ and $T_{A} = 300 K$

Substituting these values in $(i)$ we get

$n = \frac{P_{A} V_{A}}{R T_{A}} = \frac{10^{5} \times 10^{- 4}}{25 / 3 \times 300} = 0.004 mol$

Thus, at point $B$ , gas temperature: $\frac{P_{A}}{T_{A}} = \frac{P_{B}}{T_{B}}$

or $T_{B} = \frac{P_{B}}{P_{A}} \times T_{A} = \frac{2 \times 10^{5}}{10^{5}} \times 300 = 600 K$

Similar at point $C$ , gas temperature: $\frac{T_{C}}{V_{C}} = \frac{T_{B}}{V_{B}}$

or $T_{C} = \frac{V_{C}}{V_{B}} \times T_{B} = \frac{150 \times 10^{- 6}}{100 \times 10^{- 6}} \times 600 = 900 K$

B M Sharma Solutions for Chapter: Thermodynamics, Exercise 1: ILLUSTRATIONS

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Thermodynamics, Exercise 1: ILLUSTRATIONS

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