
At $25^\circ C$ and $760mm$ of $Hg$ pressure, a gas occupies $600mL$ volume. What will be its pressure at a height where temperature is $10^\circ C$ and volume of the gas is $640mL$ ?
Answer
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Hint: We have to know that, the combined gas law is the law, which joins Gay-Lussac's law, Charles' law, and Boyle's law. It's a combination of the three recently found laws. These laws relate one thermodynamic variable to another holding all the other things steady.
Complete answer:
We have to know that, the association of the factors addresses consolidated gas law, which expresses that the proportion between the result of pressing factor volume and temperature of a framework stays consistent. We have to know that the joined gas law can be utilized to clarify the mechanics where pressing factor, temperature, and volume are influenced. For instance: forced air systems, fridges and the arrangement of mists and furthermore use in liquid mechanics and thermodynamics.
Consolidated gas law can be numerically communicated and the two substances are looked at in two changed conditions, the law can be expressed as,
$\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}$
Where,
The initial pressure ${P_1} = 760mmHg$
The final pressure ${P_2} = ?$
The initial volume ${V_1} = 600mL$
The final volume ${V_2} = 640mL$
The temperature is converted to kelvin.
The initial temperature ${T_1} = 25^\circ C = 298K$
The final temperature ${T_2} = 10^\circ C = 283K$
Then applying all the given expression in the combined gas law expression,
$\dfrac{{\left( {760mmHg} \right) \times \left( {600mL} \right)}}{{\left( {298K} \right)}} = \dfrac{{\left( {{P_2}} \right) \times \left( {640mL} \right)}}{{\left( {283K} \right)}}$
Then, rewrite the above expression to calculate the final pressure.
${P_2} = \dfrac{{\left( {760mmHg} \right) \times \left( {600mL} \right) \times \left( {283K} \right)}}{{\left( {298K} \right) \times \left( {640mL} \right)}}$
Therefore,
The final pressure ${P_2} = 676.635mmHg$ .
Note:
We have to know that, the ideal gas law relates the four autonomous actual properties of a gas whenever. The best gas law can be utilized in stoichiometry issues in which synthetic responses include gases. Standard temperature and pressing factors are a valuable arrangement of benchmark conditions to think about different properties of gases.
Complete answer:
We have to know that, the association of the factors addresses consolidated gas law, which expresses that the proportion between the result of pressing factor volume and temperature of a framework stays consistent. We have to know that the joined gas law can be utilized to clarify the mechanics where pressing factor, temperature, and volume are influenced. For instance: forced air systems, fridges and the arrangement of mists and furthermore use in liquid mechanics and thermodynamics.
Consolidated gas law can be numerically communicated and the two substances are looked at in two changed conditions, the law can be expressed as,
$\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}$
Where,
The initial pressure ${P_1} = 760mmHg$
The final pressure ${P_2} = ?$
The initial volume ${V_1} = 600mL$
The final volume ${V_2} = 640mL$
The temperature is converted to kelvin.
The initial temperature ${T_1} = 25^\circ C = 298K$
The final temperature ${T_2} = 10^\circ C = 283K$
Then applying all the given expression in the combined gas law expression,
$\dfrac{{\left( {760mmHg} \right) \times \left( {600mL} \right)}}{{\left( {298K} \right)}} = \dfrac{{\left( {{P_2}} \right) \times \left( {640mL} \right)}}{{\left( {283K} \right)}}$
Then, rewrite the above expression to calculate the final pressure.
${P_2} = \dfrac{{\left( {760mmHg} \right) \times \left( {600mL} \right) \times \left( {283K} \right)}}{{\left( {298K} \right) \times \left( {640mL} \right)}}$
Therefore,
The final pressure ${P_2} = 676.635mmHg$ .
Note:
We have to know that, the ideal gas law relates the four autonomous actual properties of a gas whenever. The best gas law can be utilized in stoichiometry issues in which synthetic responses include gases. Standard temperature and pressing factors are a valuable arrangement of benchmark conditions to think about different properties of gases.
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