Aquantity of n2 gas originally held at 5.23 atm pressure in a 1.20 −l container at 26 ∘c is transferred to a 14.5 −l container at 20 ∘c. a quantity of o2 gas originally at 5.21 atm and 26 ∘c in a 5.10 −l container is transferred to this same container.

n (N₂) =  0.256 mol

n (O₂) = 1.0848 mol

n (Total) = 1.3408 mol

Pressure in new Container =  2.222 atm

Explanation:

Data Given:

For Nitrogen gas (N₂)

Pressure of N₂ gas =  5.23 atm

Volume of N₂ gas = 1.20 L

Temperature of N₂ gas = 26° C

Temperature of N₂ gas in Kelven (K) = 26° C +273

Temperature of N₂ gas in Kelven (K) = 299K

ideal gas Constant R = 0.08206 L atm K⁻¹ mol⁻¹

quantity of gas N₂ gas = ?

For Oxygen gas (O₂)

Pressure of O₂ gas =  5.21 atm

Volume of O₂ gas = 5.21 L

Temperature of O₂ gas = 26° C

Temperature of O₂ gas in Kelven (K) = 26° C +273

Temperature of O₂ gas in Kelven (K) = 299K

ideal gas Constant R = 0.08206 L atm K⁻¹ mol⁻¹

quantity of gas O₂ gas = ?

*we also have to find the total Pressure in the new container = ?

Formula Used

                         PV =nRT

                        n (N₂) = PV /RT . . . . . . . . . . . . . (1)

* Find the quantity of N₂

Put value in formula (1)

                n (N₂) = 5.23 atm x 1.20 L / 0.08206 L atm K⁻¹ mol⁻¹ x 299K

                n (N₂) =  6.276 atm .L /  24.52 L atm. mol⁻¹ x 299K

                n (N₂) =  0.256 mol

* Find the quantity of O₂

Put value in formula (1)

                n (O₂) = 5.21 atm x 5.10 L / 0.08206 L atm K⁻¹ mol⁻¹ x 299K

                n (O₂) =  26.6 atm .L /  24.52 L atm. mol⁻¹

                n (O₂) = 1.0848 mol

*Now to find the Total Quantity of both gases

                n(Total) =  n (N₂) + n (O₂)

                 n (Total) = 0.256 mol + 1.0848 mol

                 n (Total) = 1.3408 mol

**To find the Total Pressure in the new Container

Data to calculate Total Pressure in new container

Volume of gas = 14.5 L

Temperature of gases = 20° C

Temperature of gases in Kelven (K) = 20° C +273

Temperature of gases in Kelven (K) = 293K

ideal gas Constant R = 0.08206 L atm K⁻¹ mol⁻¹

Volume Pressure in new container = ?

Formula Used

                         PV =nRT

                        P = nRT / V . . . . . . . . . . . . . (2)

Put values in Equation (2)

           P =  1.3408 mol x 0.08206 L atm K⁻¹ mol⁻¹ x 293 K / 14.5 L

           P =  2.222 atm

i think it is d. e because the next one is igneous.