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A bulb of three litre capacity filled with air is heated from \small 27^{\circ}C to \small t^{\circ}C. The air thus expelled measured 1.45L at \small 17^{\circ}C. Considering the pressure to be 1 atm throughout the experiment and ignoring the expansion of bulb, calculate temprature (in oC).

Option: 1


Option: 2


Option: 3


Option: 4


Answers (1)


We have:
Initial volume of bulb = 3L
Initial temperature = 273 + 27 = 300K
Final temperature = 273 + t
Thus, according to charles’ law, we have:

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

Now at 17^{\circ}C, the volume of gas is 1.45L, thus, we have:

\mathrm{\frac{1.45}{290} =\frac{V}{t\: +\:273}}...............................(i)

Now the gas is heated from 27^{\circ}C to t^{\circ}C, thus volume can be given as follows:

\mathrm{\frac{3}{300}\: =\: \frac{3\, +\, V}{273\, +\, V}\: =\: \frac{V}{273\, +\, t}\: +\: \frac{3}{273\, +\, t}}

Using equation (i) we have:

\mathrm{\frac{3}{300}\: =\: \frac{3\, +\, V}{273\, +\, V}\: =\: \frac{1.45}{290}\: +\: \frac{3}{273\, +\, t}}

Thus, t=327^{\circ}C

Posted by

Ritika Kankaria

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