Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

How do you calculate the final pressure of a gas that is compressed from a volume of \[\;20.0{\text{ }}d{m^3}\] to \[{10.0^{}}\;d{m^3}\]and cooled from \[{100^ \circ }c{\text{ }}to{\text{ }}{25^ \circ }c\] if the initial pressure is 1 bar?

seo-qna
Last updated date: 20th Jun 2024
Total views: 373.8k
Views today: 8.73k
Answer
VerifiedVerified
373.8k+ views
Hint: The combined gas law looks at the conduct of a steady measure of gas when pressure, volume and additionally temperature is permitted to change.
The most straightforward numerical equation for the combined gas law is:
$K = \dfrac{{PV}}{T}$
In words, the result of pressure duplicated by volume and separated by temperature is a constant.
Be that as it may, the law is typically used to look at previously/after conditions. The combined gas law is communicated as:
\[\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}\]
where:
\[\begin{array}{*{20}{l}}
  {{P_i} = {\text{ }}initial{\text{ }}pressure} \\
  {{V_i} = {\text{ }}initial{\text{ }}volume} \\
  {{T_i} = {\text{ }}initial{\text{ }}absolute{\text{ }}temperature} \\
  {{P_f} = {\text{ }}final{\text{ }}pressure} \\
  {{V_f} = {\text{ }}final{\text{ }}volume} \\
  {{T_f} = {\text{ }}final{\text{ }}absolute{\text{ }}temperature}
\end{array}\]

Complete step by step answer:
Your instrument of decision here will be the joined gas law condition, which resembles this
\[\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}\]
Here
\[{P_1},{\text{ }}{V_1},{\text{ }}and{\text{ }}{T_1}\] speak to the pressure , volume, and temperature of the gas at an underlying state
\[{P_2},{\text{ }}{V_2},{\text{ }}and{\text{ }}{T_2}\] speak to the pressure, volume, and temperature of the gas at a last state
Prior to doing whatever else, ensure that you convert the temperatures from degrees Celsius to Kelvin by utilizing the way that
\[T\left[ {{\text{ }}K{\text{ }}} \right]{\text{ }} = {\text{ }}t\left[ {^ \circ C{\text{ }}} \right]{\text{ }} + {\text{ }}273.15\]
Presently, the thought here is that decreasing the volume of the gas will make the pressure increment. Then again, decreasing the temperature of the gas will make its pressure decrease.
You would thus be able to state that the adjustment in volume and the adjustment in temperature will "complete" one another, for example whichever change is more significant will decide whether the pressure increases or decreases.
Along these lines, improve the consolidated gas law to settle for \[{P_2}\]
 \[\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}\]
${P_2} = \dfrac{{{V_1}}}{{{V_2}}} \times \dfrac{{{T_2}}}{{{T_1}}} \times {P_1}$
Plug in your qualities to discover
${{\text{P}}_2} = 2 \times 0.799 \times 1{\text{bar}}$
 \[ = 1.6bar\]

Note: I'll leave the appropriate response adjusted to two sig figs, however remember that you just have one significant figure for the underlying pressure of the gas.
As should be obvious, the decrease in volume was more significant than the decrease in temperature; accordingly, the pressure of the gas increased.