Question

It is desired to increase the volume of 80 $c{m}^3$ of a gas by 20% without changing pressure. To what temperature the gas be heated if its initial temperature is ${25}^{\circ }C$?

• A. 83.4C.
• B. 86.7C.
• C. 93.4C.
• D. 102.3C.

Answer 1

The correct option is A

83.4C.

The desired increase in the volume of gas is
20% of 80 $c{m}^3$ = $\frac{20}{100}\times 80$ = 16 $c{m}^3$

Thus, the final volume of the gas is 80 + 16 = 96 $c{m}^3$

Now, V₁ = 80 $c{m}^3$ , V₂ =96 $c{m}^3$

T₁ = 25${}^{\circ }$C =(273+25)K = 298 K, T₂ = ?

Applying Charles's law, we get
$\frac{V_1}{T_1}=\frac{V_2}{T_2}$
$\frac{80}{298}=\frac{96}{T_2}$
T₂ = 357.6 K = 357.6 - 273 = 83.4${}^{\circ }$C.