Question

An unknown gas has a pressure of 699.0 mm Hg at 40.0 C. What is the temperature at standard pressure?

Answer 1

Hello!

An unknown gas has a pressure of 699.0 mm Hg at 40.0 C. What is the temperature at standard pressure ?

We have the following information:

P1 (initial pressure) = 699 mmHg

T1 (initial temperature) = 40 ºC (in Kelvin)

TK = TºC + 273.15 → TK = 40 + 273.15 → T1 (initial temperature) = 313.15 K

P2 (final pressure) = 1 atm (in STP) → P2 (final pressure) = 760 mmHg

T2 (final temperature) = ? (in Kelvin)

According to the Law of Charles and Gay-Lussac in the study of gases, we have an isochoric (or isovolumetric) transformation when its volume remains constant or equal, then we will have the following formula:

$\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$

$\dfrac{699}{313.15} = \dfrac{760}{T_2}$

$699*T_2 = 760*313.15$

$699\:T_2 = 237994$

$T_2 = \dfrac{237994}{699}$

$T_2 = 340.4778255... \to \boxed{\boxed{T_2 \approx 340\:K}}\end{array}}\qquad\checkmark$

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