Question

A gas at 89°C occupies a volume of 0.67 L. At what Celsius temperature will the volume increase to 1.12 L?

Answer 1

Answer:

V1 / T1 = V2 / T2

0.67 / (273 + 89) = 1.12 / (273 + t )

t = 332 °C

Answer 2

The temperature will be 332°C.

Given values:

Volume,

• $V_1 = 0.67 \ L$
• $V_2 = 1.12 \ L$

Temperature,

• $T_1 = 89^{\circ} C \ or \ 362 \ K$
• $T_2 = \ ?$

We know the relation,

→ $\frac{V_1}{T_1} = \frac{V_2}{T_2}$

By substituting the values, we get

→ $\frac{0.67}{362} = \frac{1.12}{T_2}$

→   $T_2 = \frac{1.12\times 362}{0.67}$

          $= 332^{\circ} C$

Thus the answer above is right.

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