A quantity of gas is collected in a graduated tube over mercury. The volume of the gas at 20°C is 50.0 mL and the level of the mercury in the tube is 100 mm above the outside mercury level. The barometer reads 750 mm. Volume of the gas at STP is

Question

A quantity of gas is collected in a graduated tube over mercury. The volume of the gas at 20°C is 50.0 mL and the level of the mercury in the tube is 100 mm above the outside mercury level. The barometer reads 750 mm. Volume of the gas at STP is (A) 40 mL (B) 42 mL (C) 39.8 mL (D) 60 mL

Tardigrade Admin · Accepted Answer

T1 = 20 + 273 = 293 K, T2 = 273 K V1 = 50 mL. P1 = 750-100 = 650 mm Hg; P2 = 760 mm Hg (P1V1/T1) = (P2 V2/T2) V2 = (P1V1T2/T1P2) = (50 × 650 × 273/293 × 760) = 39.8 mL

Q. A quantity of gas is collected in a graduated tube over mercury. The volume of the gas at 20 $^{\circ}$ C is 50.0 mL and the level of the mercury in the tube is 100 mm above the outside mercury level. The barometer reads 750 mm. Volume of the gas at STP is

40 mL

42 mL

39.8 mL

60 mL

$T_{1}$ = 20 + 273 = 293 K, $T_{2}$ = 273 K
$V_{1}$ = 50 mL.
$P_{1}$ = 750-100 = 650 mm Hg; $P_{2}$ = 760 mm Hg
$\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$
$V_{2} = \frac{P _{1} V _{1} T _{2}}{T _{1} P _{2}} = \frac{50 \times 650 \times 273}{293 \times 760}$
= 39.8 mL

Q. A quantity of gas is collected in a graduated tube over mercury. The volume of the gas at 20∘C is 50.0 mL and the level of the mercury in the tube is 100 mm above the outside mercury level. The barometer reads 750 mm. Volume of the gas at STP is