20.0 dm³ of an ideal gas ‘X’ at 600 K and 0.5 MPa undergoes isothermal reversible expansion until pressure of the gas is 0.2 MPa. Which of the following option is correct? (Given: log 2 = 0.3010 and log 5 = 0.6989)

20.0 dm3 of an ideal gas at 600 K and 0.5 MPa undergoes isothermal reversible expansion until pressure becomes 0.2 MPa

Q. 20.0 dm³ of an ideal gas ‘X’ at 600 K and 0.5 MPa undergoes isothermal reversible expansion until pressure of the gas is 0.2 MPa. Which of the following option is correct?

(Given: log 2 = 0.3010 and log 5 = 0.6989)

(A) w = −9.1 kJ, ΔU = 0, ΔH = 0, q = +9.1 kJ

(B) w = 9.1 J, ΔU = 9.1 J, ΔH = 0, q = 0

(D) w = +4.1 kJ, ΔU = 0, ΔH = 0, q = −4.1 kJ

Correct Answer: (A)

Explanation

The process is isothermal and reversible for an ideal gas.

For an isothermal process of an ideal gas:

ΔU = 0 and ΔH = 0

Work done in reversible isothermal expansion is given by:

w = −2.303 nRT log (P₁ / P₂)

Using ideal gas equation to find nRT:

nRT = P₁V₁

= 0.5 × 10⁶ Pa × 20 × 10⁻³ m³

= 10,000 J

Now,

w = −2.303 × 10,000 × log (0.5 / 0.2)

log (0.5 / 0.2) = log 2.5 = log 5 − log 2 = 0.6989 − 0.3010 = 0.3979

w = −2.303 × 10,000 × 0.3979 ≈ −9.1 × 10³ J

w ≈ −9.1 kJ

Since ΔU = 0, from first law of thermodynamics:

q = −w = +9.1 kJ

Thus, the correct option is (A).

Explanation

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