Question #cc924

1 Answer
Apr 13, 2017

The volume of the dry gas at STP is #"49 cm"^3#.

Explanation:

The gas is saturated with water vapour, so

#p_text(gas) + p_text(water) = p_text(atm)#, or

#p_text(gas) = p_text(atm) - p_text(water) = "765 mmHg - 5 mmHg" = "760 mmHg"#

We can use the Combined Gas Laws to calculate the volume at STP.

#color(blue)(|bar(ul(color(white)(a/a)(p_1V_1)/T_1 = (p_2V_2)/T_2color(white)(a/a)|)))" "#

We can rearrange this formula to get

#V_2 = V_1 × p_1/p_2 × T_2/T_1 #

STP is defined as 1 bar and 0 °C.

#V_1 = "50 cm"^3#
#p_1 = 760 color(red)(cancel(color(black)("mmHg"))) × (1 color(red)(cancel(color(black)("atm"))))/(760 color(red)(cancel(color(black)("mmHg")))) × (101.325 color(red)(cancel(color(black)("kPa"))))/(1 color(red)(cancel(color(black)("atm")))) × "1 bar"/(100 color(red)(cancel(color(black)("kPa")))) = "1.01 bar"#
#p_2 = "1 bar"#
#T_2 =color(white)(ll) "0 °C" = "273.15 K"#
#T_1 = "10 °C" = "283.15 K"#

#V_2 = "50 cm"^3 × (1.01 color(red)(cancel(color(black)("bar"))))/(1 color(red)(cancel(color(black)("bar")))) × (273.15 color(red)(cancel(color(black)("K"))))/(283.15 color(red)(cancel(color(black)("K")))) = "49 cm"^3#