Question

It is desired to increase the volume of the 800 ml of a gas by 20% keeping the pressure constant . To what temperature the gas be heated , if the initial temperature is 22 ° C.

Answer 1

Answer:-

Given:

Initial Volume (V1) = 800 ml

And,

Volume is increased by 20% with increase of temperature.

→ Final Volume (V2) = 800 * (120) / 100 = 960 ml

Initial Temperature (T2) = 22° C = 273 + 22 = 295° K

Let,

Final Temperature (T1) = T K.

We know that,

Charle's law of gases states that at constant pressure the Volume of a gas is directly proportional to it's temperature.

That is,

V ∝ T

→ V/T = k (Constant)

$\sf \implies \: \dfrac {V _{1}}{T _{1}} = \dfrac{V _{2}}{T_{2} } \\ \\ \implies \sf \dfrac{ 800 }{295^{\circ}} = \dfrac{960 }{T } \\ \\ \sf \implies 800\times T= 295\times 960 \\ \\ \implies \sf T = \dfrac { 295 \times 960} {800} \\ \\ \implies \sf \large \red {{\: T= 354^{\circ} K}}$

→ T = 354° - 273°

→ T = 81° C

Therefore, the temperature of the gas increases to 81° C when the Volume of the gas is increased by 20%.

Answer 2

Given:

• Initial Volume= V1= 800 ml
• Volume is increased by 20%
• Initial temperature= 22°C = 22+ 273K = 295K

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Increase in Volume

=20%

=$800 \times \dfrac{120}{100}$

= 160 ml

Therefore Final volume = V2

=800+160ml

=960ml

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Considering the gas as an ideal gas. Apply Charles law.

According to Charles Law at constant pressure the volume of a ideal gas is directly proportional to it's temperature.

The formula is:

$\dfrac{v1}{t1} = \dfrac{v2}{t2}$

$==> \dfrac{800}{295} = \dfrac{960}{t2}$

$\implies \: T2 = \dfrac{960 \times 295}{800}$

$\red{\bold{ \implies \: T2 = 354k}}$

Also; 354K

= 354-273 °C

=$\red{\bold{81°C}}$

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Therefore the gas can be heated till T2= 354K= 81°C