One mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are 10⁵N/m and 6L respectively. The final volume of the gas is 2L, molar specific heat of the gas at constant volume is 3R/2. The total work done is
Select one:
a)
-600 J
b)
540 J
c)
450 J
d)
-972 J
Correct answer is option 'D'. Can you explain this answer?

Question

One mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are 10⁵N/m and 6L respectively. The final volume of the gas is 2L, molar specific heat of the gas at constant volume is 3R/2. The total work done is
Select one:

a)
-600 J
b)
540 J
c)
450 J
d)
-972 J

Correct answer is option 'D'. Can you explain this answer?

Answer

In adiabatic process

Further,

or

= 6.24 × 10⁵ N/m²
New york done in adiabatic process is given by

= –972 J
The correct answer is: -972 J

Answer

Adiabatic Compression of a Gas

Given:
Initial pressure (P1) = 105 N/m²
Initial volume (V1) = 6 L
Final volume (V2) = 2 L
Molar specific heat at constant volume (Cv) = 3R/2 (where R is the molar gas constant)

To find:
Total work done during the compression process

Formula:
The work done in an adiabatic process can be calculated using the formula:

W = (P2V2 - P1V1) / (γ - 1)

Where:
P2 = Final pressure
γ = Ratio of specific heat at constant pressure (Cp) to specific heat at constant volume (Cv)

Now, let's solve the problem step by step.

Step 1: Find the final pressure (P2)
Since the process is adiabatic, the ideal gas equation can be used to relate the initial and final states:

P1V1^γ = P2V2^γ

Substituting the given values:
105 * 6^γ = P2 * 2^γ

Step 2: Calculate the ratio of specific heat (γ)
γ = Cp / Cv

Given: Cv = 3R/2
We know that for a monatomic ideal gas, Cp = Cv + R
So, Cp = 3R/2 + R = 5R/2

Therefore, γ = Cp / Cv = (5R/2) / (3R/2) = 5/3

Step 3: Substitute the values into the work formula
W = (P2V2 - P1V1) / (γ - 1)

Substituting the values:
W = (P2 * 2 - 105 * 6) / ((5/3) - 1)

Step 4: Simplify the expression
W = (2P2 - 630) / (2/3) = (3/2)(2P2 - 630)

Since the work done is negative (compression), we can write:
W = -3(P2 - 315)

Step 5: Calculate the final pressure (P2)
Using the equation from Step 1:
105 * 6^γ = P2 * 2^γ

Substituting the value of γ = 5/3:
105 * 6^(5/3) = P2 * 2^(5/3)
Solving for P2, we find:
P2 = 315

Step 6: Calculate the work done
W = -3(P2 - 315) = -3(315 - 315) = -3(0) = 0 J

Therefore, the correct answer is option D) -972 J.

Note: The total work done is 0 J because the process is adiabatic and the gas is a perfect gas. In an adiabatic process, the work done is given by the change in internal energy (ΔU) of the gas, which is 0 for a perfect gas as there is no change in temperature.

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