Question

A gas is cooled from 365 K to 285 K while its volume changes from 12.8 L to 9.9 L. The initial pressure of the gas is 1.9 atm. The final pressure of the gas, rounded to the nearest tenth, is atm.

Answer 1

The correct answer to the question is : 1.9 atm.

EXPLANATION:

Let us consider the pressure, volume and temperature of a gas are denoted as P,V and T respectively.

Initial condition :                                                   Final condition:

$P_{1} = 1.9\ atm$                                      $P_{2} =\ ?$

$V_{1} =\ 12.8\ L$                                       $V_{2} =\ 9.9\ L$

$T_{1}=\ 365\ K$                                        $T_{2} =\ 285\ K$

From combined equation of gas, we know that-

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

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

                                 $=\ \frac{1.9\times 12.8}{365}\times \frac{285}{9.9}\ atm$

                                 $=\ 1.91814\ atm$

                                 $=\ 1.9\ atm$              [ans]

Hence, the final pressure is 1.9 atm.

Answer 2

A gas is cooled from 365 K to 285 K while its volume changes from 12.8 L to 9.9 L. The initial pressure of the gas is 1.9 atm. The final pressure of the gas, rounded to the nearest tenth, is atm.