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.

Respuesta :

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.

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:

[tex]P_{1} = 1.9\ atm[/tex]                                      [tex]P_{2} =\ ?[/tex]

[tex]V_{1} =\ 12.8\ L[/tex]                                       [tex]V_{2} =\ 9.9\ L[/tex]

[tex]T_{1}=\ 365\ K[/tex]                                        [tex]T_{2} =\ 285\ K[/tex]

From combined equation of gas, we know that-

                           [tex]\frac{P_{1}V_{1}}{T_{1} } =\ \frac{P_{2}V_{2}}{T_{2}}[/tex]

                        ⇒ [tex]P_{2} =\ \frac{P_{1}V_{1}}{T_{1} } \times \frac{T_{2} }{V_{2} }[/tex]

                                 [tex]=\ \frac{1.9\times 12.8}{365}\times \frac{285}{9.9}\ atm[/tex]

                                 [tex]=\ 1.91814\ atm[/tex]

                                 [tex]=\ 1.9\ atm[/tex]              [ans]

Hence, the final pressure is 1.9 atm.

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