Question

18. 400 mL of oxygen are collected over water at
25° C and 0.948 bar pressure. Find the volume of
dry gas at S.T.P. (Aqueous tension at
,25° C = 0.0318 bar). 33547 ml​

Answer 1

Answer : The volume of dry gas at STP is, 340.50 ml

Explanation :

First we have to calculate the pressure of dry gas.

$\text{Pressure of dry gas}=0.948-0.0318=0.9162bar$

Now we have to calculate the volume of dry gas at STP.

Using the combined gas equation :

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas (dry gas) = 0.9162 bar

$P_2$ = final pressure of gas = 0.986 bar

$V_1$ = initial volume of gas = 400 ml

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $25^oC=273+25=298K$

$T_2$ = final temperature of gas (at STP)  = $0^oC=273+0=273K$

Now put all the given values in the above equation, we get the final volume of gas.

$\frac{0.9162bar\times 400ml}{298K}=\frac{0.986bar\times V_2}{273K}$

$V_2=340.50ml$

Therefore, the volume of dry gas at STP is, 340.50 ml

Answer 2

Explanation:

First we have to calculate the pressure of dry gas.

\text{Pressure of dry gas}=0.948-0.0318=0.9162barPressure of dry gas=0.948−0.0318=0.9162bar

Now we have to calculate the volume of dry gas at STP.

Using the combined gas equation :

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}T1P1V1=T2P2V2

where,

P_1P1 = initial pressure of gas (dry gas) = 0.9162 bar

P_2P2 = final pressure of gas = 0.986 bar

V_1V1 = initial volume of gas = 400 ml

V_2V2 = final volume of gas = ?

T_1T1 = initial temperature of gas = 25^oC=273+25=298K25oC=273+25=298K

T_2T2 = final temperature of gas (at STP)  = 0^oC=273+0=273K0oC=273+0=273K

Now put all the given values in the above equation, we get the final volume of gas.

\frac{0.9162bar\times 400ml}{298K}=\frac{0.986bar\times V_2}{273K}298K0.9162bar×400ml=273K0.986bar×V2

V_2=340.50mlV2=340.50ml

Therefore, the volume of dry gas at STP is, 340.50 ml

hope it will help you . plese mark it s a brilliant ans