Answer:
2.57832 atm
Explanation:
[tex]P_1[/tex] = Initial pressure = 2.1 atm
[tex]P_2[/tex] = Fianl pressure
[tex]T_1[/tex] = Initial Temperature = (21+273.15) K
[tex]T_2[/tex] = Final Temperature = (88+273.15) K
From Gay-Lussacs law we have
[tex]\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\\\Rightarrow P_2=\dfrac{P_1\times T_2}{T_1}\\\Rightarrow P_2=\dfrac{2.1\times (88+273.15)}{21+273.15}\\\Rightarrow P_2=2.57832\ atm[/tex]
The final pressure would be 2.57832 atm
Answer:
2.58 atm
Explanation:
P1 = 2.1 atm
T1 = 21 °C = 294 K
T2 = 88 °C = 361 K
P2 = ?
By using the gas laws
P / T = constant keeping the volume constant.
P1 / T1 = P2 / T2
2.10 / 294 = P2 / 361
P2 = 2.58 atm
Thus, the pressure becomes 2.58 atm.
