A gas mixture contain l gm. H2 and 1 gm. He if temp. of gas mixture increase 0°C to 100°C at isobaric process. Than find given heat of gas mixture (γHe = 5/3, γH2 = 7/5, R = 2 cal/molek)

To find the heat given to the gas mixture during an isobaric process when the temperature increases from 0°C to 100°C, we need to consider the specific heat capacities of the gases involved and the total number of moles in the mixture. Let's break this down step by step.

Step 1: Calculate the number of moles of each gas

First, we need to determine the number of moles of hydrogen (H2) and helium (He) in the mixture. The molar mass of H2 is approximately 2 g/mol, and for He, it is about 4 g/mol.

• For H2:

  Number of moles (n_H2) = mass / molar mass = l g / 2 g/mol = 0.5 moles

• For He:

  Number of moles (n_He) = mass / molar mass = 1 g / 4 g/mol = 0.25 moles

Step 2: Determine the total number of moles in the mixture

The total number of moles (n_total) in the gas mixture is the sum of the moles of H2 and He:

n_total = n_H2 + n_He = 0.5 + 0.25 = 0.75 moles

Step 3: Calculate the specific heat at constant pressure (C_p)

For an ideal gas, the specific heat at constant pressure can be calculated using the formula:

C_p = (γ * R) / (γ - 1)

Now, we will calculate C_p for both gases:

• For H2 (γ_H2 = 7/5):

  C_p,H2 = (7/5 * 2 cal/mole·K) / (7/5 - 1) = (14/5) / (2/5) = 7 cal/mole·K

• For He (γ_He = 5/3):

  C_p,He = (5/3 * 2 cal/mole·K) / (5/3 - 1) = (10/3) / (2/3) = 5 cal/mole·K

Step 4: Calculate the average specific heat capacity of the mixture

The average specific heat capacity (C_p,mix) of the mixture can be calculated using the mole fractions of each gas:

C_p,mix = (n_H2 * C_p,H2 + n_He * C_p,He) / n_total

Substituting the values:

C_p,mix = (0.5 * 7 + 0.25 * 5) / 0.75 = (3.5 + 1.25) / 0.75 = 4.75 / 0.75 = 6.33 cal/mole·K

Step 5: Calculate the heat added to the gas mixture

Using the formula for heat transfer at constant pressure:

Q = n_total * C_p,mix * ΔT

Where ΔT is the change in temperature (100°C - 0°C = 100 K):

Q = 0.75 moles * 6.33 cal/mole·K * 100 K = 474.75 cal

Final Result

The total heat given to the gas mixture during the isobaric process when the temperature increases from 0°C to 100°C is approximately 474.75 calories.