An open container of volume 3 litre contains air at 1 atmospheric pressure- The container is heated from initial temperature 27 -C to t -C- the amount of the gas expelled from the container measured 1-45 litre at 17-C and 1 atm- Find temperature -t-

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Byju's Answer

Standard X

Chemistry

Charles' Law

An open conta...

Question

An open container of volume 3 litre contains air at 1 atmospheric pressure. The container is heated from initial temperature $27^{\circ} C$ to $t^{\circ} C$ , the amount of the gas expelled from the container measured 1.45 litre at $17^{\circ} C$ and 1 atm. Find temperature $^{'} t^{'}$ .

327^{\circ} C

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316^{\circ} C

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336^{\circ} C

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307^{\circ} C

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Solution

The correct option is A $327^{\circ} C$
According to Charle's law,
$V \propto T$ at constant pressure
Hence increase in temperature, increases the volume. Hence to maintain constant pressure, some volume of the gas is expelled.

Initial volume of the gas $V_{1} = 3 L$
Initial temperature of the gas $T_{1} = 27^{\circ} C = 300 K$
Final volume $V_{2} = initial volume + volume expelled$
$= 3 + Δ V L$
Final temperature $T_{2} = ?$
Determination of $Δ V$ at $T_{2}$ :
Given $Δ V L$ at $17^{\circ} C$ or 290 K = 1.45 L

Accoridng to Charle's law at constant pressure $V_{1} T_{2} = V_{2} T_{1}$
$Δ V L$ at temperature $T_{2}$
$Δ V$ = $\frac{1.45 \times T_{2}}{290}$

Hence final volume $V_{2}$ = 3 + $\frac{1.45 \times T_{2}}{290}$
Again applying Charle's law,

$∴ T_{2} = \frac{V_{2} T_{1}}{V_{1}}$
$= \frac{(3 + \frac{1.45 \times T_{2}}{290}) \times 300}{3} = 600 K$

$∴ T_{2} = 600 K$ = $327^{\circ} C$

Theory:

Charles Law :

For a fixed amount of gas at constant pressure, volume occupied by the gas is directly proportional to absolute temperature of the gas.

or

It states "at constant pressure, the volume of a given mass of a gas, increases or decreases by

$\frac{1}{273.15}$ th of its volume at $0^{o} C$ for every rise or fall of one degree in temperature.

From charles law :

$P r e s s u r e (P) = c o n s t a n t$

$N u m b e r o f m o l e s o f g a s = c o n s t a n t$

For a fixed amount of gas at constant pressure :

$V \propto T$ or $V = k_{2} T$

$\frac{V}{T} = k_{2} c o n s t a n t$

Here $k_{2}$ constant depends on the amount of the gas and pressure of the gas.

Applying charles law for two different states of a given amount of gas at constant pressure :

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$

Where :

$V_{1} = I n i t i a l V o l u m e$ , $V_{2} = F i n a l V o l u m e$ , $P_{1} = I n i t i a l P r e s s u r e$ and $P_{2} = F i n a l P r e s s u r e$

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An open container of volume 3 litre contains air at 1 atmospheric pressure. The container is heated from initial temperature 27 ∘C to t ∘C, the amount of the gas expelled from the container measured 1.45 litre at 17∘C and 1 atm. Find temperature ′t′.

An open container of volume 3 litre contains air at 1 atmospheric pressure. The container is heated from initial temperature $27^{\circ} C$ to $t^{\circ} C$ , the amount of the gas expelled from the container measured 1.45 litre at $17^{\circ} C$ and 1 atm. Find temperature $^{'} t^{'}$ .