Question

(20) Four moles of a gas expand isothermally at 300 K.
If the final pressure of the gas is 80% of the initial
pressure, find the work done by the gas on its
surroundings. (R = 8.314 J/mol-K) (2 marks)​

Answer 1

Answer

• Work done by gas = -2228.7 Joules

Explanation

Given:-

• 4 moles of gas expand isothermally, n = 4 moles
• Constant temperature for expansion, T = 300 K
• Final pressure of the gas (P₂) = 80 % of initial gas pressure (P₁)
• Universal gas constant, R = 8.314 J/mol-K

To find:-

• Work done by the gas, W =?

Formula required:-

• Formula for the work done in an isothermal process

       W = 2.303 nRT log₁₀(P₂/P₁)

[ Where, W is work done by gas, n is number of moles of gas, R is universal gas constant, T is constant temperature, P₂ is final pressure and P₁ is initial pressure ]

Solution:-

Given that,

→ Final temperature (P₂) = 80 % of initial temperature (P₁)

→ P₂ = 80/100 × P₁

→ P₂/P₁ = 4/5

                     

Now, Using the formula for work done in an isothermal process

→ W = 2.303 nRT log₁₀(P₂/P₁)

→ W = 2.303 × 4 × 8.314 × 300 × log₁₀(4/5)

→ W = 2.303 × 4 × 8.314 × 300 × -0.097

→ W = -2228.7  J

Therefore,

Work done by the gas is -2228.7 Joules.

Answer 2

Given:

No. of moles of gas (n) = 4

Temperature of expansion (T) = 300 K

Final pressure (P₂) = 80% of initial pressure (P₁)

Universal gas constant (R) = 8.314 J/mol-K

To find:

Work done by the gas.

Solution:

Work done in an isothermal expansion,

(W) = 2.303nRTlog₁₀(V₂/V₁)

Since,

P₁V₁ = P₂V₂

V₂/V₁ = P₂/P₁

(W) = 2.303nRTlog₁₀(P₂/P₁)

_______________________

P₂ = 80% P₁

P₂ = 80/100 x P₁

P₂ = 4/5P₁

Dividing both sides by P₁ :-

P₂P₁= 4/5P₁ x 1/P₁

P₂P₁ = 4/5

_______________________

Substituting values in the formula:

W = 2.303nRTlog₁₀(P₂/P₁)

W = 2.303 × 4 × 8.314 × 300 × log₁₀(4/5)

W = 22,976.57 × -0.097

W = - 2228.7 J