120 g of an ideal gas of molecular weight 40 g mol -1 are confined to a volume of 20 L at 400 K. Using R=0.0821 L atm K-1 mol -1, the pressure of the gas is

Question

120 g of an ideal gas of molecular weight 40 g mol -1 are confined to a volume of 20 L at 400 K. Using R=0.0821 L atm K-1 mol -1, the pressure of the gas is (A) 4.90 atm (B) 4.92 atm (C) 5.02 atm (D) 4.96 atm

Tardigrade Admin · Accepted Answer

P V=R T, P V=(W/M) R T 20 P=(120/40) × .0821 × 400 or P=4.92 atm

Q. $120 g$ of an ideal gas of molecular weight $40 g m o l^{- 1}$ are confined to a volume of $20 L$ at $400 K$ . Using $R = 0.0821 L$ atm $K^{- 1} m o l^{- 1}$ , the pressure of the gas is

4.90 atm

4.92 atm

5.02 atm

4.96 atm

$P V = RT,$
$P V = \frac{W}{M} RT$
$20 P = \frac{120}{40} \times .0821 \times 400$
or $P = 4.92 a t m$

Q. 120g of an ideal gas of molecular weight 40gmol−1 are confined to a volume of 20L at 400K. Using R=0.0821L atm K−1mol−1, the pressure of the gas is