An ideal gas at 77- C is compressed adiabatically and reversibly to 9-16 of its original volume- If Y -5-3- then the rise in temperature is-Given- 1-331-33-1-46A- 161 KB- 450 KC- 375 KD- 313 K

The correct option is A 161 KFor an adiabatic process nbsp; T_2T_1 = (V_1V_2)^γ 1 where, T_1= initial temperature T_2= final nbsp;temperature V_1= initi ...

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Standard XII

Chemistry

First Law of Thermodynamics

An ideal gas ...

Question

An ideal gas at $77^{o} C$ is compressed adiabatically and reversibly to $\frac{9}{16}$ of its original volume. If $γ = \frac{5}{3},$ then the rise in temperature is:
Given: ${1.33}^{1.33} = 1.46$

450 K

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375 K

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313 K

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161 K

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Solution

The correct option is D $161 K$
For an adiabatic process
$\frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1}$
where, $T_{1} =$ initial temperature
$T_{2} =$ final temperature
$V_{1} =$ initial volume
$V_{2} =$ final volume $= \frac{9}{16} V_{1}$
$γ = \frac{5}{3}$
$\Rightarrow T_{2} = 350 {(\frac{16}{9})}^{\frac{5}{3} - 1} = 350 {(\frac{16}{9})}^{\frac{2}{3}}$
$= 350 {(\frac{4}{3})}^{\frac{4}{3}} = 511 K$
therefore rise in tempearture is :
$\Rightarrow △ T = 513 - 350 = 161 K$

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First Law of Thermodynamics

Standard XII Chemistry

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An ideal gas at 77oC is compressed adiabatically and reversibly to 916 of its original volume. If γ=53, then the rise in temperature is: Given: 1.331.33=1.46

The correct option is D 161 KFor an adiabatic process T2T1=(V1V2)γ−1 where, T1= initial temperature T2= final temperature V1= initial volume V2= final volume =916V1 γ=53 ⇒T2=350(169)53−1=350(169)23 =350(43)43=511K therefore rise in tempearture is : ⇒△T=513−350=161 K

An ideal gas at $77^{o} C$ is compressed adiabatically and reversibly to $\frac{9}{16}$ of its original volume. If $γ = \frac{5}{3},$ then the rise in temperature is:
Given: ${1.33}^{1.33} = 1.46$