Question

A vessel of 50 L contains 40 g of an ideal gas X at 600 K. The pressure exerted by the gas is 1 atm. 20 g of an
ideal gas Y is added to the same vessel keeping the same temperature. Total pressure becomes 3 atm. If
the molar mass of gas X is m (g/mol) and the molar mass of gas Y is n (g/mol) then find the value of (m+ni).
(Use R=
1 atm L
12 mol K​

Answer 1

Answer: 60 is the answer (m+n) that is 50+10

Explanation:

Answer 2

Answer:

The value of (m+n) is 50 g/mole.

Explanation:

For an ideal gas,

PV = nRT

Substituting the values in the above expression

1 × 50 = n × 1/12 ×600

n         = (50 × 12)/600

          = 1 mole

1 mole is equivalent to 40 g of X gas.

After adding 20 g of an ideal gas

Given: total pressure = 3 atm

Pressure exerted by Y alone = 3-1

                                                = 2 atm

Substituting the values in the expression

2 × 50 = n × 1/12 ×600

n         = (50 × 12)/600

          = 2 moles

20 gram of Y gas is equivalent to 2 moles.

Therefore, 1 mole = 10 g of Y gas

The value of (m+n) = 40+10

                               = 50 g/mole

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