A volume of gas starts at a pressure of 10 atmospheres (atm) and a temperature of 27°C. If the temperature is increased by 75°C and the volume of the gas remains constant, what is the new pressure? About the author Hadley
Given that, Initial pressure = 10 atm Initial temperature = 27 °C Final temperature = 75 °C We have to find final pressure of the gas[tex].[/tex] ❖ As per Gay Lussac’s law at constant volume, pressure of an ideal gas is directly proportional to the temperature. Mathematically, P ∝ T First of all we have to convert the temperature into kelvin. ⟩ Initial temperature = 27 °C = 300 K ⟩ Final temperature = 75 °C = 348 K By substituting the given values; [tex]\sf:\implies\:\dfrac{P_1}{P_2}=\dfrac{T_1}{T_2}[/tex] [tex]\sf:\implies\:\dfrac{10}{P_2}=\dfrac{300}{348}[/tex] [tex]\sf:\implies\:P_2=\dfrac{10\times 348}{300}[/tex] [tex]\sf:\implies\:P_2=\dfrac{348}{30}[/tex] [tex]:\implies\:\underline{\boxed{\bf{\gray{P_2=11.6\:atm}}}}[/tex] Reply
Given that,
Initial pressure = 10 atm
Initial temperature = 27 °C
Final temperature = 75 °C
We have to find final pressure of the gas[tex].[/tex]
❖ As per Gay Lussac’s law at constant volume, pressure of an ideal gas is directly proportional to the temperature.
Mathematically, P ∝ T
First of all we have to convert the temperature into kelvin.
⟩ Initial temperature = 27 °C = 300 K
⟩ Final temperature = 75 °C = 348 K
By substituting the given values;
[tex]\sf:\implies\:\dfrac{P_1}{P_2}=\dfrac{T_1}{T_2}[/tex]
[tex]\sf:\implies\:\dfrac{10}{P_2}=\dfrac{300}{348}[/tex]
[tex]\sf:\implies\:P_2=\dfrac{10\times 348}{300}[/tex]
[tex]\sf:\implies\:P_2=\dfrac{348}{30}[/tex]
[tex]:\implies\:\underline{\boxed{\bf{\gray{P_2=11.6\:atm}}}}[/tex]