Question

please answer this question 13

Answer 1

Answer:- 32.8 mL

Solution:- The problem is based on combined gas law equation, the equation is written as:

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

Initial volume of the gas, $V_1$ = 38.0 mL

initial temperature, $T_1$ = 27 + 273 = 300 K

the gas is moist means it's stored by water displacement and so the total given pressure is the sum of partial pressures of nitrogen gas and water vapors. hence, we need to subtract the water vapor pressure from total pressure to get the partial pressure of the nitrogen gas:

partial pressure of nitrogen gas, $P_1$ = 746.5mm - 26.5mm = 720mm

$T_2$ = 0+273 = 273 K

$P_2$ = 760mm

$V_2$ = ?

Let's plug in the values in the equation and solve it for final volume:

$\frac{720mm(38.0mL)}{300K}=\frac{760mm(V_2)}{273K}$

on rearranging this for final volume:

$V_2=\frac{720mm(38.0mL)(273K)}{760mm(300K)}$

$V_2$ = 32.8 mL

So, the volume of the nitrogen gas at 0 degree C and 760 mm of pressure is 32.8 mL.