You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Gay Lussac's Law, Avagadro's Law

120 g of an i...

Question

120 g of an ideal gas of molecular weight 40 is confined to a volume of 20 litres at 400 K, then the pressure of a gas is:

490 atm

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4.92 atm

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2236 atm

No worries! We‘ve got your back. Try BYJU‘S free classes today!

22.4 atm

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 4.92 atm
We know that,

PV =nRT ....ideal gas equation

$P = \frac{3 0.082 \times 400}{20}$

$P = 4.92 a t m$

Hence, the correct option is $B$

Suggest Corrections

Similar questions

Q. 120

g

of an ideal gas of molecular weight 40 are confined to a volume of 20 litre at 400

K

, then the pressure of gas is................(in

a t m

as nearest integer)

Q. The molecular weight of a gas is

40

. At

400

K if

120

g of this gas has a volume of

20

litres, the pressure of the gas in atm is:

Q. 4

g

of an ideal gas occupies 5.6035 litres of volume at 546

K

and 2

a t m

pressure. What is its molecular weight?

Q. The pressure exerted by

12 g

of an ideal gas at temperature

1^{o} C

in a vessel of volume

V

litre is one atm. When the temperature is increased by

10

degrees at the same volume, the pressure increases by

10

%. Calculate the temperature

t

and volume

V

. (Molecular weight of the gas

= 120

)

Q. 20 litres of a monoatomic ideal gas at

0^{o} C

and 20

a t m

pressure is suddenly exposed to 1

a t m

pressure and the gas expands adiabatically against this constant pressure to maximum possible volume. The final temperature and volume of the gas respectively are

Join BYJU'S Learning Program

120 g of an ideal gas of molecular weight 40 is confined to a volume of 20 litres at 400 K, then the pressure of a gas is:

The correct option is B 4.92 atmWe know that,PV =nRT ....ideal gas equationP=30.082×40020P=4.92atmHence, the correct option is B

The correct option is B 4.92 atm
We know that,

PV =nRT ....ideal gas equation

$P = \frac{3 0.082 \times 400}{20}$

$P = 4.92 a t m$

Hence, the correct option is $B$