An ideal gas at 27o C is compressed adiabatically and reversibly to 827 of its original volume- If - - 53- then the rise in temperature is

For an adiabatic process nbsp; T_2T_1 = (V_1V_2)^γ 1 where, T_1= initial temperature T_2= final nbsp;temperature V_1= initial volume V_2= final volume =82 ...

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Standard XII

Chemistry

Work Done in Reversible Adiabatic Process

An ideal gas ...

Question

An ideal gas at $27^{o} C$ is compressed adiabatically and reversibly to $\frac{8}{27}$ of its original volume. If $γ = \frac{5}{3},$ then the rise in temperature is

450 K

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375 K

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225 K

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405 K

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Solution

The correct option is B $375 K$
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 $= \frac{8}{27} V_{1}$
$γ = \frac{5}{3}$
$\Rightarrow T_{2} = 300 {(\frac{27}{8})}^{\frac{5}{3} - 1} = 300 {(\frac{27}{8})}^{\frac{2}{3}}$
$= 300 {(\frac{3}{2})}^{2} = 300 (\frac{9}{4}) = 675 K$
therefore rise in tempearture is :
$\Rightarrow △ T = 675 - 300 = 375 K$

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Work Done in Reversible Adiabatic Process

Standard XII Chemistry

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An ideal gas at 27oC is compressed adiabatically and reversibly to 827 of its original volume. If γ=53, then the rise in temperature is

The correct option is B 375 K For an adiabatic process T2T1=(V1V2)γ−1 where, T1= initial temperature T2= final temperature V1= initial volume V2= final volume =827V1 γ=53 ⇒T2=300(278)53−1=300(278)23 =300(32)2=300(94)=675K therefore rise in tempearture is : ⇒△T=675−300=375 K

An ideal gas at $27^{o} C$ is compressed adiabatically and reversibly to $\frac{8}{27}$ of its original volume. If $γ = \frac{5}{3},$ then the rise in temperature is