Answer:
The volume of the gas sample at standard pressure is 819.5ml
Explanation:
Solution Given:
let volume be V and temperature be T and pressure be P.
[tex] V_1=250ml[/tex]
[tex] V_2=?[/tex]
[tex] P_{total}=735 mmhg[/tex]
1 torr= 1 mmhg
42.2 torr=42.2 mmhg
so,
[tex] P_{water}=42.2mmhg[/tex]
[tex] T_1=35°C=35+273=308 K[/tex]
Now
firstly we need to find the pressure due to gas along by subtracting the vapor pressure of water.
[tex] P_{gas}=P_{total}-P_{water} [/tex]
=735-42.2=692.8 mmhg
Now
By using combined gas law equation:
[tex]\frac{P_1*V_1}{T_1} =\frac{P_2*V_2}{T_2}[/tex]
[tex]V_2=\frac{P_1*}{P_2}*\frac{T_2}{T_1} *V_1[/tex]
[tex]V_2=\frac{P_gas}{P_2}*\frac{T_2}{T_1} *V_1[/tex]
Here [tex]P_2 \:and\: T_2[/tex] are standard pressure and temperature respectively.
we have
[tex]P_2=750mmhg \:and\: T_2=273K[/tex]
Substituting value, we get
[tex]V_2=\frac{692.8}{750}*\frac{273}{308} *250[/tex]
[tex]V_2= 819.51 ml[/tex]