A certain amount of an ideal gas is initially at a pressure P1 and temperature T1- First- it undergoes an isobaric process 1 - 2 such that T2 - 3T14- Then- it undergoes an isochoric process 2 - 3 such that T3 - T22- The ratio of the final volume to the initial volume of the ideal gas is

Given that nbsp; P_1 = Initial pressure T_1 = Initial Temperature nbsp; V_3V_1 = ?? Process 1 2 is constant pressure process nbsp; T_2 = 3T_14 (Given) ...

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Standard XIII

Physics

Ideal Gas Equation

A certain amo...

Question

A certain amount of an ideal gas is initially at a pressure $P_{1}$ and temperature $T_{1}$ . First, it undergoes an isobaric process 1 - 2 such that $T_{2} = \frac{3 T_{1}}{4}$ . Then, it undergoes an isochoric process $2 \to 3$ such that $T_{3} = \frac{T_{2}}{2}$ . The ratio of the final volume to the initial volume of the ideal gas is

0.25

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1

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1.5

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0.75

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Solution

The correct option is D $0.75$

Given that
$P_{1}$ = Initial pressure
$T_{1}$ = Initial Temperature
$\frac{V_{3}}{V_{1}} = ? ?$
Process 1 - 2 is constant pressure process
$T_{2} = \frac{3 T_{1}}{4}$ (Given) $\Rightarrow \frac{T_{2}}{T_{1}} = \frac{3}{4}$
For constant pressure in ideal gas,
$V \propto T$ (Charle's law)
$\Rightarrow \frac{V_{2}}{V_{1}} = \frac{T_{2}}{T_{1}} = \frac{3}{4} - - - (1)$
Process 2 -3 is constant volume process.
$\frac{T_{2}}{2} = T_{3}$ (Given) $\Rightarrow \frac{T_{3}}{T_{2}} = \frac{1}{2}$
Now,
$\frac{V_{3}}{V_{1}} = \frac{final volume}{Initial volume} = \frac{V_{2}}{V_{1}}$
$[∵ V_{3} = V_{2}]$
$= \frac{3}{4} = 0.75$ [from eq (1)]

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Ideal Gas Equation

Standard XIII Physics

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A certain amount of an ideal gas is initially at a pressure P1 and temperature T1. First, it undergoes an isobaric process 1 - 2 such that T2=3T14. Then, it undergoes an isochoric process 2→3 such that T3=T22. The ratio of the final volume to the initial volume of the ideal gas is