Thermodynamics

Step 1: Process I – Adiabatic Reversible Process:

For an adiabatic reversible process, the first law of thermodynamics tells us that the change in internal energy is equal to the work done by the system since no heat is exchanged with the surroundings.
Given the equation , we can calculate the change in internal energy:
.
Since the process is adiabatic, the work done is equal to the change in internal energy:
.

Step 2: Process II – Reversible Isothermal Process:

In the second process, the temperature is given as 900 K.
Since the process is isothermal, the temperature remains constant during the process.
We are tasked with finding the temperature after the reversible adiabatic process.

Step 3: Calculation of after the Reversible Adiabatic Process:

For an adiabatic process, the relation between the change in temperature and internal energy is given by:
,
where:
is the number of moles, is the specific heat capacity at constant volume, is the initial temperature, and is the final temperature.
For helium, we know that . Now, substitute the given values into the equation:
.
Canceling out from both sides and solving for :
.
Substitute (assuming one mole of helium), and simplify the equation:
.
Multiplying both sides by 2:
.
Now, divide by 5:
.
Finally, solving for :
.

Step 4: Work Done in Process II:

The work done in a reversible isothermal process is given by:
,
where:
and .
Substitute these values into the equation:
.
Thus, the work done in process II is:
.

Final Answer:
The correct answer is: .

Given Data:

The molar heat capacity of the gas at constant pressure is:

Solution:

We are asked to find the value of , where is the volume at state . The process involves two parts: (adiabatic) and (isobaric). We can relate the total heat absorbed during these processes using the formula for heat in an ideal gas:

The total heat absorbed is , so we need to use this information to find the relationship between and the other variables in the process. The calculations will show that the value of equals 7.

Final Answer:

The value of is .

Calculation of at 500 K

Given: The equation for the equilibrium constant is related to the forward and backward rate constants and as:

Thus, we can express as:

Step 1: Solving for

We are given that , and also that at 500 K. Therefore, we can substitute these values into the equation:

Solving for :

Step 2: Arrhenius Equation

The Arrhenius equation gives the rate constant as:

Taking the natural logarithm of both sides:

Multiplying both sides by 2.303 to convert to base 10 logarithms:

Step 3: Solving for

Now, we use the equation for the logarithmic difference of the rate constants at two temperatures and :

Simplifying this expression:

Step 4: Finding

Next, we calculate the logarithmic ratio of the rate constants at 250 K and 500 K:

This simplifies to:

Step 5: Final Calculation

Solving for , we get:

Taking the absolute value, we find:

Conclusion

The correct answer is:

5

Calculation of Mass of Product S

Given: 4-methyloct-1-ene (P, 2.52 g) reacts with HBr in the presence of (C6H5CO)2O2, producing two isomeric bromides in a 9:1 ratio, with a combined yield of 50%. The primary alkyl bromide reacts with diethylamine and is treated with aqueous K2CO3 to give a non-ionic product S in 100% yield.

Step 1: Calculate the moles of 4-methyloct-1-ene (P)

The molecular weight of 4-methyloct-1-ene (C8H15) is:

The moles of 4-methyloct-1-ene (P) are:

Step 2: Moles of Primary Alkyl Bromide

The two isomeric bromides are in a 9:1 ratio, and the combined yield is 50%. Thus, the moles of the primary alkyl bromide are:

Step 3: Reaction with Diethylamine

The primary alkyl bromide reacts with diethylamine to give product S in 100% yield, so the moles of product S are:

Step 4: Molar Mass of Product S

The molar mass of product S is the sum of the mass of the alkyl group (4-methyloct-1-ene) plus the mass of the diethylamino group:

Step 5: Mass of Product S

The mass of product S is given by:

Step 6: Final Calculation

The mass of product S obtained is 100% of the yield, so we multiply the result by 1000 to convert to milligrams:

Conclusion

The final mass of product S obtained is:

1791 mg

Calculation of at 250K

Given the equation for the equilibrium constant at 500 K:

Since

We have the equation:

Rearranging it:

Solving for :

The relationship between and is given by:

Using the Arrhenius equation for :

Taking the natural logarithm of both sides:

Multiplying by to convert to log base 10:

Now, solving for :

Thus:

Next, using the relationship for and :

Substitute the values for and :

Solving this gives:

Finally, the value of is:

The absolute value is .

Final Answer:

The correct answer is .

Combined yield = 50% 
50% of 2.52 = 1.26 g 
M = 126 
n4-methyloct-1-ene = 0.01 
90% of 0.01 = 0.009 
Mass of S = 0.009 × 199 
= 1.791 g 
= 1791 mg.

Thermodynamic Calculation for Ideal Monoatomic Gas

We are given the following information for one mole of an ideal monoatomic gas undergoing two reversible processes:

• Process A → B: This is an adiabatic process.
• Process B → C: This process is isothermal.
• Total heat absorbed in both processes:
• Molar heat capacity at constant pressure:

Step 1: Heat Exchange in Adiabatic Process (A → B)

For an adiabatic process, no heat is exchanged. Hence, the heat exchange in process is zero:

Step 2: Heat Exchange in Isothermal Process (B → C)

For the isothermal process , the temperature is constant, and the heat absorbed is equal to the work done by the gas. The heat absorbed in this process is given by the equation:

The work done during an isothermal expansion is given by the equation:

So, the heat absorbed during process is:

Step 3: Total Heat Absorbed

The total heat absorbed in the entire process (both and ) is:

The problem states that the total heat absorbed is , so we equate:

Dividing both sides by , we get:

Therefore, we have:

Step 4: Finding

We need to find the value of . Using the equation , we get:

Taking the logarithm of both sides:

Now, multiplying by 2:

Since is unknown, but the value of is asked, we assume that is negligible and proceed with the simplified result:

Conclusion

The value of is .

Step 1: Write the balanced equation for electrolysis of water Step 2: Moles of water Step 3: Moles of gaseous products Step 4: Expansion work done