The question is in the attached picture. I know how to solve the first part of the problem, but the second part confused me a lot.

Now when the oven (system) is unsealed, doesn’t make that an open system where ideal gas equation is not applicable?

If not, then would the new pressure and temperature be STD since it’s open system now? Which parameter is constant in this case and which is variable? I’m totally confused.

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    $\begingroup$ As you asked this on Academia, a comment suggested you ask your professor as you indicated you are a TA... $\endgroup$ – Solar Mike Feb 5 '19 at 21:40
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    $\begingroup$ You may find Kreuzer and Payne's "Thermodynamics of heating a room" interesting/useful. $\endgroup$ – Chemomechanics Feb 5 '19 at 23:18
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    $\begingroup$ I'm voting to close this question as off-topic because First, this is a crosspost from Academia, second because this is Physics, not Engineering. $\endgroup$ – Carl Witthoft Feb 6 '19 at 15:52
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    $\begingroup$ Shows no effort either... $\endgroup$ – Solar Mike Feb 7 '19 at 6:40
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    $\begingroup$ It is engineering or chemistry ... not physics. Gas laws are applied in general chemistry, in much of chemical engineering processing of gases, and sometimes one might find also in mechanical engineering. $\endgroup$ – Jeffrey J Weimer Feb 18 '19 at 23:13

For the first part, with an ideal gas in a closed system

$$ V = \mathrm{constant} \ \Rightarrow \ \frac{p_i}{T_i} = \frac{p_f}{T_f} $$

For the second part, with an ideal gas in an open system

$$ p = \mathrm{constant} \ \Rightarrow \ \frac{V_i}{T_i} = \frac{V_f}{T_f} = \frac{V_i + \Delta V}{T_f} $$

Both cases are one equation with one unknown.

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