Verified Answer :

dexteright02

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:

\(\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}\)

\(\dfrac{699}{313.15} = \dfrac{760}{T_2}\)

\(699*T_2 = 760*313.15\)

\(699\:T_2 = 237994\)

\(T_2 = \dfrac{237994}{699}\)

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

_______________________________

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

The temperature of the gas at STP has been 340.5 K.

The gas has been assumed to be an ideal gas. From the variations of the ideal gas equation:

\(\rm \dfrac{Initial\;Pressure}{Final\;Temperature}\;=\;\dfrac{Final\;Pressure}{Final\;Temperature}\)

The initial/known pressure of gas = 699 mm Hg

The initial temperature = 40\(\rm ^\circ C\) = 313.15 K

At STP, the standard pressure has been = 760 mm Hg.

The temperature at STP can be calculated as:

\(\rm \dfrac{699}{313.15}\;=\;\dfrac{760}{Final\;Temperature}\)

2.232 = \(\rm \dfrac{760}{Final\;Temperature}\)

Final temperature = \(\rm \dfrac{760}{2.232}\) K

Final temperature = 340.50 K.

The temperature of the gas at STP has been 340.5 K.

For more information about the temperature of the gas, refer to the link:

https://brainly.com/question/15354399

Related Questions