Question

if the initial temperature of an ideal gas at 2.250 atm is 62.00°c, what final temperature would cause the pressure to be reduced to 1.700 atm? t = °c

Explanation:

Step1: Convert initial temperature to Kelvin

The formula to convert Celsius to Kelvin is \(T(K)=T(^{\circ}C)+273.15\).
For \(T_1 = 62.00^{\circ}C\), \(T_1=62.00 + 273.15=335.15\space K\)

Step2: Apply Gay - Lussac's Law

Gay - Lussac's Law is \(\frac{P_1}{T_1}=\frac{P_2}{T_2}\), where \(P_1 = 2.250\space atm\), \(T_1 = 335.15\space K\), \(P_2=1.700\space atm\) and \(T_2\) is the final temperature in Kelvin.
We can solve for \(T_2\): \(T_2=\frac{P_2T_1}{P_1}\)
Substitute the values: \(T_2=\frac{1.700\times335.15}{2.250}\)
\(T_2=\frac{569.755}{2.250}\approx253.22\space K\)

Step3: Convert final temperature to Celsius

Use \(T(^{\circ}C)=T(K)-273.15\)
\(T_2 = 253.22- 273.15=-19.93^{\circ}C\)

Answer:

\(-19.93^{\circ}C\)