Answer:
[tex]P_2=20atm[/tex]
Explanation:
Hello,
In this case, we apply the Gay-Lussac's law which allows us to understand the pressure-temperature behavior as a directly proportional relationship:
[tex]\frac{P_1}{T_1}= \frac{P_2}{T_2}[/tex]
Thus, we solve for the final pressure P2 to obtain it as shown below:
[tex]P_2=\frac{P_1T_2}{T_1}=\frac{10atm*20K}{10K} \\\\P_2=20atm[/tex]
Hence, we notice that the temperature doubles as well as the pressure.
Best regards.
