Question

Exactly 5.00 L of air at -50.0°C is warmed to 100.0°C. What is the new volume if the pressure remains constant?

Answer 1

Explanation:

The given data is as follows.

   $V_{1}$ = 5.00 L,                           $V_{2}$ = ?

   $T_{1}$ = -50.0 + 273 = 223 K,     $T_{2}$ = 100.0 + 273 = 373 K

     P = constant

Therefore, calculate $V_{2}$ as follows.

      $\frac{PV_{1}}{T_{1}}$ = $\frac{PV_{2}}{T_{2}}$

Since pressure is constant so, P will be cancelled out from both the sides.

      $\frac{V_{1}}{T_{1}}$ = $\frac{V_{2}}{T_{2}}$

            $V_{2}$ = $\frac{V_{1}T_{2}}{T_{1}}$

                                     = $\frac{5.00 L \times 373 K}{223 K}$

                                     = $\frac{1865 L}{223}$

                                     = 8.36 L

Thus, we can conclude that $V_{2}$ is 8.36 L.