A 2.5 L Of Nitrogen Gas Exerts A Pressure Of 760 Mmhg At 473k.What Temperature Is Needed To Reduce The Volume To 1.75l At 1140 Torr?

A 2.5 L Of Nitrogen Gas Exerts A Pressure Of 760 Mmhg At 473k.What Temperature Is Needed To Reduce The Volume To 1.75l At 1140 Torr?

a 2.5 L of nitrogen gas exerts a pressure of 760 mmhg at 473k.what temperature is needed to reduce the volume to 1.75L at 1140 torr?

Hello!

A 2.5 L of nitrogen gas exerts a pressure of 760 mmhg at 473k.what temperature is needed to reduce the volume to 1.75L at 1140 torr ?

We have the following information:

V1 (initial volume) = 2.5 L

V2 (final volume) = 1.75 L

T1 (initial temperature) = 473 K

T2 (final temperature) = ? (in Kelvin)

P1 (initial pressure) = 760 mmHg → P1 (initial pressure) = 1 atm

P2 (final pressure) = 1140 torr (in atm)

1 atm --- 760 torr

x atm ---- 1140 torr

760x = 1140

x = 1140/760

x = 1.5 atm → P2 (final pressure) = 1.5 atm

***Note: In SI the pressure can be mmHg or atm

Now, we apply the data of the variables above to the General Equation of Gases, lets see:

$\dfrac{P_1*V_1}{T_1} =\dfrac{P_2*V_2}{T_2}$

$\dfrac{1*2.5}{473} =\dfrac{1.5*1.75}{T_2}$

$\dfrac{2.5}{473} =\dfrac{2.625}{T_2}$

multiply the means by the extremes

$2.5*T_2 = 473*2.625$

$2.5\:T_2 = 1241.625$

$T_2 = \dfrac{1241.625}{2.5}$

$\boxed{\boxed{T_2 = 496.65\:K}}\Longleftarrow(final\:temperature)\end{array}}\qquad\checkmark$

Answer:

The temperature is 496.65 Kelvin

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I Hope this helps, greetings ... Dexteright02! =)

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