Question

Consider 4.60 L of a gas at 365 mmHg and 20 C . If the container is compressed to 2.60 L and the temperature is increased to 36 C , what is the new pressure, P2, inside the container? Assume no change in the amount of gas inside the cylinder.

Answer 1

Explanation:

The given data is as follows.

       $V_{1}$ = 4.60 L,        $P_{1}$ = 365 mm Hg

       $V_{2}$ = 2.60 L,       $P_{2}$ = ?

       $T_{1}$ = (20 + 273) K = 293 K,      $T_{2}$ = (36 + 273) K = 309 K

Since, number of moles of gas are equal so, according to ideal gas equation:

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

          $\frac{365 mm Hg \times 4.60 L}{293 K}$ = $\frac{P_{2} \times 2.60 L}{309 K}$

          $P_{2}$ = 681.03 mm Hg

Thus, we can conclude that new pressure $P_{2}$ is 681.03 mm Hg.