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Work Done in Isothermal Reversible Process

5.6 litre of ...

Question

5.6 litre of helium gas at STP is adiabatically compressed to $0.7$ litre. Taking the initial temperature to be $T_{1}$ , the work done on the gas is:

\frac{9}{8} R T_{1}

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\frac{3}{2} R T_{1}

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\frac{15}{8} R T_{1}

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\frac{9}{8} R T_{1}

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Solution

The correct option is A $\frac{9}{8} R T_{1}$
For an adiabatic process, $T V^{γ - 1} = c o n s t a n t$
or $T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$

where $V_{1} = 5.6 L$ and $V_{2} = 0.7 L$
also, $H e$ is monoatomic so $γ = \frac{5}{3}$
so, $T_{1} (5.6)^{\frac{2}{3}} = T_{2} (0.7)^{\frac{2}{3}}$
this gives $T_{2} = 4 T_{1}$

$V_{1} = 5.6 L$ at STP so, moles of gas $n = \frac{5.6}{22.4} = \frac{1}{4}$

work done by the gas in adiabatic process is given by,
$w = \frac{P_{2} V_{2} - P_{1} V_{1}}{γ - 1}$
$= \frac{n R T_{2} - n R T_{1}}{γ - 1}$
$= \frac{n R Δ T}{γ - 1}$

Substituting these values to get
$w = \frac{R (4 T_{1} - T_{1})}{4 \times 2 / 3}$
$w = \frac{9}{8} R T_{1}$

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Work Done in Isothermal Reversible Process

Standard XII Chemistry

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5.6 litre of helium gas at STP is adiabatically compressed to 0.7 litre. Taking the initial temperature to be T1, the work done on the gas is:

5.6 litre of helium gas at STP is adiabatically compressed to $0.7$ litre. Taking the initial temperature to be $T_{1}$ , the work done on the gas is: