1 mole of NH3 gas at 27 oC is expanded in reversible adiabatic condition to make volume 64 times of its initial volume- - - 43 The magnitude of work done in cal will be - Give your answer in cal and upto two decimal only

For a reversible adiabatic process : T_1V_1^γ 1 = nbsp;T_2V_2^γ 1 Given, T_1 =300 K V_2=64V_1 putting the values we get, ⇒ 300 × V^1/3_1 = T_2(64V_1)^1/3 ...

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Standard XI

Chemistry

Work Done in Reversible Adiabatic Process

1 mole of NH3...

Question

1 mole of $N H_{3}$ gas at $27^{o} C$ is expanded in reversible adiabatic condition to make volume 64 times of its initial volume. $(γ = \frac{4}{3})$ The magnitude of work done (in cal) will be :
(Give your answer in cal and upto two decimal only)

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Solution

For a reversible adiabatic process :
$T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$
Given, $T_{1} = 300 K$
$V_{2} = 64 V_{1}$
putting the values we get,
$\Rightarrow 300 \times V_{1}^{1 / 3} = T_{2} (64 V_{1})^{1 / 3} T_{2} = 75 K$
So, final temperature is $75 K$

and work done will be ,
$W = n C_{v} (T_{2} - T_{1})$
we know $C_{v} = \frac{R}{γ - 1}$
so,
$W = 1 \times \frac{R}{γ - 1} (75 - 300) = \frac{2}{\frac{4}{3} - 1} (- 225) = - 1350 c a l$

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1 mole of NH3 gas at 27 oC is expanded in reversible adiabatic condition to make volume 64 times of its initial volume. (γ=43) The magnitude of work done (in cal) will be : (Give your answer in cal and upto two decimal only)

For a reversible adiabatic process : T1Vγ−11=T2Vγ−12 Given, T1 =300 K V2=64V1 putting the values we get, ⇒300×V1/31=T2(64V1)1/3T2=75 K So, final temperature is 75 K and work done will be , W=nCv(T2−T1) we know Cv=Rγ−1 so, W=1×Rγ−1(75−300)=243−1(−225)=−1350 cal

1 mole of $N H_{3}$ gas at $27^{o} C$ is expanded in reversible adiabatic condition to make volume 64 times of its initial volume. $(γ = \frac{4}{3})$ The magnitude of work done (in cal) will be :
(Give your answer in cal and upto two decimal only)