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Byju's Answer

Standard XII

Chemistry

Liquefaction of Gases and Critical Constants

An ideal diat...

Question

An ideal diatomic gas is caused to pass through a cycle shown on the $P - V$ diagram in figure, where $V_{2} = 3.00$ Litres. If $P_{1}, V_{1}$ and $T_{1}$ specify the state 1, then the temperature of the state 3 is:

(T_{1} / 3)^{1.4}

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T_{1} / 3^{1.4}

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T_{1} / 3^{0.4}

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cannot be determined

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Solution

The correct option is D cannot be determined
R.E.F image
Given :
$V_{2} = 3 l t$
State 1 $\to$ 2 is isothermal,
so, $T_{1} = T_{2}$ .
State 1 $\to$ 3 is Adiabatic,
$T V^{γ - 1} = c o n s t$
$T_{1} V_{1}^{γ - 1} = T_{3} V_{3}^{γ - 1}$
$T_{3} = T_{1} {(\frac{V_{1}}{V_{3}})}^{γ - 1}$
$V_{3} = V_{2}$ {Isometric}
$T_{3} = T_{1} {(\frac{V_{1}}{V_{2}})}^{γ - 1}$
$T_{3} = T_{1} {(\frac{V_{1}}{3})}^{γ - 1}$
value of $V_{1}$ is not given.
So, $T_{3}$ can not be determined

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An ideal diatomic gas is caused to pass through a cycle shown on the P−V diagram in figure, where V2=3.00 Litres. If P1,V1 and T1 specify the state 1, then the temperature of the state 3 is:

The correct option is D cannot be determinedR.E.F imageGiven : V2=3ltState 1 → 2 is isothermal,so, T1=T2.State 1→3 is Adiabatic,TVγ−1=constT1Vγ−11=T3Vγ−13T3=T1(V1V3)γ−1V3=V2 {Isometric}T3=T1(V1V2)γ−1T3=T1(V13)γ−1value of V1 is not given.So, T3 can not be determined

An ideal diatomic gas is caused to pass through a cycle shown on the $P - V$ diagram in figure, where $V_{2} = 3.00$ Litres. If $P_{1}, V_{1}$ and $T_{1}$ specify the state 1, then the temperature of the state 3 is: