Answer:
2. [tex]V_2=17L[/tex]
3. [tex]V=82.9L[/tex]
Explanation:
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2. In this case, we can evidence the problem by which volume and temperature are involved, so the Charles' law is applied to:
[tex]\frac{V_2}{T_2}=\frac{V_1}{T_1}[/tex]
Thus, considering the temperatures in kelvins and solving for the final volume, V2, we obtain:
[tex]V_2=\frac{V_1T_2}{T_1}[/tex]
Therefore, we plug in the given data to obtain:
[tex]V_2=\frac{18.2L(22+273)K}{(45+273)K} \\\\V_2=17L[/tex]
3. In this case, it is possible to realize that the 3.7 moles of neon gas are at 273 K and 1 atm according to the STP conditions; in such a way, considering the ideal gas law (PV=nRT), we can solve for the volume as shown below:
[tex]V=\frac{nRT}{P}[/tex]
Therefore, we plug in the data to obtain:
[tex]V=\frac{3.7mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm}\\\\V=82.9L[/tex]
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