Question

An ideal gas in a sealed container has an initial volume of 2.80 L. At constant pressure, it is cooled to 18.00 °C, where its

Answer 1

Answer:

$T_1=-91.18\°C$

Explanation:

Hello there!

In this case, given the T-V variation, we understand it is possible to apply the Charles' law as shown below:

$\frac{T_1}{V_1}= \frac{T_2}{V_2}$

Thus, since we are interested in the initial temperature, we can solve for T1, plug in the volumes and use T2 in kelvins:

$T_1= \frac{T_2V_1}{V_2}\\\\T_1= \frac{(18.00+273.15)K(1.75L)}{(2.80L)}\\\\T_1=182K-273.15\\\\T_1=-91.18\°C$

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