Verified Answer :

Willchemistry

V₁ = 42.3 mL

T₁ ( K ) = 98.15ºC + 273 = 371.15 K

V₂ = ?

T₂ ( K) = -18.50ºC + 273 = 254.5 K

V₁ / T₁  = V₂ / T₂

42.3 / 371.15 =  V₂ / 254.5

371.15 x V₂ = 42.3 x  254.5

371.15 x V₂ = 10765.35

V₂ = 10765.35 / 371.15

V₂ = 29.00 mL

hope this helps!

Cricetus

The volume that the gas will occupy is "29.00 mL".

According to the question,

Volume,

  • \(V_1 = 42.3 \ mL\)
  • \(V_2 = ?\)

Temperature,

  • \(T_1 = 98.15^{\circ} C\)

             \(= 371.15 \ K\)

  • \(T_2 = -18.50^{\circ} C\)

             \(= 254.5 \ K\)

As we know the relation,

→ \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\)

or,

→ \(V_2 = \frac{V_1\times T_2}{T_1}\)

By substituting the values, we get

→      \(= \frac{41.3\times 254.5}{371.15}\)

→      \(= \frac{10765.35}{371.15}\)

→      \(= 29.00 \ mL\)

Thus the above answer is right.  

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https://brainly.com/question/22590663

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