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Question: 28 gm of N₂ gas is contained in a flask at a pressure of 10 atm and at a temperature of 57°C. It is ...

28 gm of N₂ gas is contained in a flask at a pressure of 10 atm and at a temperature of 57°C. It is found that due to leakage in the flask, the pressure is reduced to half and the temperature reduced to 27°C. The quantity of N₂ gas that leaked out is :-

A

11/20 gm

B

5/63 gm

C

20/11 gm

D

63/5 gm

Answer

63/5 gm

Explanation

Solution

The problem involves the application of the ideal gas law for a gas in a flask under two different conditions (initial and final). The volume of the flask remains constant.

1. Identify Initial Conditions:

  • Initial mass of N₂ gas (m₁) = 28 gm
  • Molar mass of N₂ (M) = 28 gm/mol
  • Initial number of moles (n₁) = m₁ / M = 28 gm / 28 gm/mol = 1 mol
  • Initial pressure (P₁) = 10 atm
  • Initial temperature (T₁) = 57°C. Convert to Kelvin: T₁ = 57 + 273 = 330 K

2. Identify Final Conditions:

  • Final pressure (P₂) = P₁ / 2 = 10 atm / 2 = 5 atm
  • Final temperature (T₂) = 27°C. Convert to Kelvin: T₂ = 27 + 273 = 300 K
  • Let the final number of moles be n₂.
  • The volume of the flask (V) is constant.

3. Apply the Ideal Gas Law: The ideal gas law is given by PV=nRTPV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. Since the volume (V) and the gas constant (R) are constant, we can write the relationship between the initial and final states as:

P1Vn1T1=R\frac{P_1V}{n_1T_1} = R

P2Vn2T2=R\frac{P_2V}{n_2T_2} = R

Therefore,

P1n1T1=P2n2T2\frac{P_1}{n_1T_1} = \frac{P_2}{n_2T_2}

4. Calculate the Final Number of Moles (n₂): Rearranging the equation to solve for n₂:

n2=n1×P2P1×T1T2n_2 = n_1 \times \frac{P_2}{P_1} \times \frac{T_1}{T_2}

Substitute the known values:

n2=1 mol×5 atm10 atm×330 K300 Kn_2 = 1 \text{ mol} \times \frac{5 \text{ atm}}{10 \text{ atm}} \times \frac{330 \text{ K}}{300 \text{ K}}

n2=1×12×3330n_2 = 1 \times \frac{1}{2} \times \frac{33}{30}

n2=1×12×1110n_2 = 1 \times \frac{1}{2} \times \frac{11}{10}

n2=1120 moln_2 = \frac{11}{20} \text{ mol}

5. Calculate the Final Mass of N₂ Gas (m₂): The final mass of N₂ gas remaining in the flask is:

m2=n2×Mm_2 = n_2 \times M

m2=1120 mol×28 gm/molm_2 = \frac{11}{20} \text{ mol} \times 28 \text{ gm/mol}

m2=11×2820 gmm_2 = \frac{11 \times 28}{20} \text{ gm}

m2=11×75 gmm_2 = \frac{11 \times 7}{5} \text{ gm}

m2=775 gmm_2 = \frac{77}{5} \text{ gm}

6. Calculate the Quantity of N₂ Gas Leaked Out: The quantity of N₂ gas that leaked out is the difference between the initial mass and the final mass:

Mass leaked = m₁ - m₂

Mass leaked = 28 gm−775 gm28 \text{ gm} - \frac{77}{5} \text{ gm}

Mass leaked = (28×5)−775 gm\frac{(28 \times 5) - 77}{5} \text{ gm}

Mass leaked = 140−775 gm\frac{140 - 77}{5} \text{ gm}

Mass leaked = 635 gm\frac{63}{5} \text{ gm}

The quantity of N₂ gas that leaked out is 63/5 gm.