Answer:
[tex]P_2=1.90atm[/tex]
Explanation:
Hello!
In this case, according to the ideal gas equation ratio for two states:
[tex]\frac{P_1V_1}{P_2V_2} =\frac{n_1RT_1}{n_2RT_2}[/tex]
Whereas both n and R are cancelled out as they don't change, we obtain:
[tex]\frac{P_1V_1}{P_2V_2} =\frac{T_1}{T_2}[/tex]
Thus, by solving for the final pressure, we obtain:
[tex]\frac{P_2V_2}{P_1V_1} =\frac{T_2}{T_1}\\\\P_2=\frac{T_2P_1V_1}{V_2T_1}[/tex]
Now, since initial conditions are 1.00 atm, 273.15 K and 17 L and final temperature and volume are 94 + 273 = 367 K and 12 L respectively, the resulting pressure turns out to be:
[tex]P_2=\frac{367K*1.00atm*17L}{12L*273.15K}\\\\P_2=1.90atm[/tex]
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