Question

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Answer 1

Answer:

${ { { \frac{1 \frac{1 \frac{4 \frac{( \sqrt{(x} }{?} }{?} }{?} \times \frac{?}{?} \times \frac{?}{?} }{?} }^{2} }^{?} \times \frac{?}{?} }^{2}$

Answer 2

Answer:

Given:

Initial Volume (V1) = 800 ml = 800/1000 = 4/5 L

And,

Volume is increased by 20% with increase of temperature.

→ Final Volume (V2) = 800 * (120) / 100 = 960 ml = 960/1000 = 24/25 L.

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

Let,

Initial 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)

\begin{gathered}\sf \implies \: \dfrac {V _{1}}{T _{1}} = \dfrac{V _{2}}{T_{2} } \\ \\ \implies \sf \dfrac{ \dfrac{4}{5} }{295^{\circ}} = \dfrac{ \dfrac{24}{25} }{T } \\ \\ \sf \implies \dfrac{4}{5} \times T= 295\times \dfrac{24}{25} \\ \\ \implies \sf T = \dfrac{5}{4} \times 295 \times \dfrac{25}{24} \\ \\ \implies \sf \large \red {{\: T= 384.11 ^{\circ} K}}\end{gathered}

⟹

T

1

V

1

=

T

2

V

2

⟹

295

∘

5

4

=

T

25

24

⟹

5

4

×T=295×

25

24

⟹T=

4

5

×295×

24

25

⟹T=384.11

∘

K

→ T = 384.11° - 273°

→ T = 111.11° C