The cylinder contains 100 gm of an ideal gas (m o l . w t . = 40 g m / m o l) at 27° C and 2 atm . pressure. In transportation the cylinder fell and a dent was created. The valve present cannot keep the pressure greater than 2 atm . Hence 10 gm of a gas got leaked out. The volume of the container before and after dent is-

Question

The cylinder contains 100 gm of an ideal gas (m o l . w t . = 40 g m / m o l) at 27° C and 2 atm . pressure. In transportation the cylinder fell and a dent was created. The valve present cannot keep the pressure greater than 2 atm . Hence 10 gm of a gas got leaked out. The volume of the container before and after dent is- (A) 30.8 L ; 27.7 L (B) 27.7 L ; 30.8 L (C) 30.8 L ; 30.8 L (D) 27.7 L ; 27.7 L

Tardigrade Admin · Accepted Answer

Before dent vol. of gas=(n R T/P) =(100/40) × (0.082 × 300/2)=30.75 L After dent vol. of gas=(n R T/P) =(90/40) × (0.082 × 300/2)=27.675 L

$30.8 L; 27.7 L$

$27.7 L; 30.8 L$

$30.8 L; 30.8 L$

$27.7 L; 27.7 L$

Before dent $v o l . o f g a s = \frac{n RT}{P}$

$= \frac{100}{40} \times \frac{0.082 \times 300}{2} = 30.75 L$

After dent $v o l . o f g a s = \frac{n RT}{P}$

$= \frac{90}{40} \times \frac{0.082 \times 300}{2} = 27.675 L$

30.8L;27.7L

27.7L;30.8L

30.8L;30.8L

27.7L;27.7L

Before dent vol.ofgas=PnRT​ =40100​×20.082×300​=30.75L After dent vol.ofgas=PnRT​ =4090​×20.082×300​=27.675L

$30.8 L; 27.7 L$

$27.7 L; 30.8 L$

$30.8 L; 30.8 L$

$27.7 L; 27.7 L$

Before dent $v o l . o f g a s = \frac{n RT}{P}$

$= \frac{100}{40} \times \frac{0.082 \times 300}{2} = 30.75 L$

After dent $v o l . o f g a s = \frac{n RT}{P}$

$= \frac{90}{40} \times \frac{0.082 \times 300}{2} = 27.675 L$