Answer:
[tex]P_2=3.41atm[/tex]
Explanation:
Hello,
In this case, we first compute the moles of oxygen in 0.654 g and 1.35 g by using its molar mass (32 g/mol):
[tex]n_1=0.654g*\frac{1mol}{32g} =0.0204mol\\n_2=1.35g*\frac{1mol}{32g} =0.0422mol[/tex]
Then, by using the ideal gas equation at the both states, given the same both temperature and volume:
[tex]V_1=V_2\\\\\frac{n_1RT}{P_1}=\frac{n_2RT}{P_2}[/tex]
[tex]\frac{n_1}{P_1}=\frac{n_2}{P_2}[/tex]
We compute the volume at the second moles:
[tex]P_2=\frac{P_1n_2}{n_1}=\frac{1.65atm*0.0422mol}{0.0204mol}\\ \\P_2=3.41atm[/tex]
Best regards.
