Question

*VERIFIED ANSWER IS NEEDED*

An ideal gas expands isothermally from
10-3 m3 to 10-2 m3 at 300 K against a constant
pressure of 105 Nm-2. The work done on the gas is


[NEET-2019 (Odisha)]

(1) - 900 kJ
(2) + 270 kJ
(3) - 900 J
(4) + 900 kJ

​

Answer 1

Correct Question:

An ideal gas expands isothermally from 10⁻³ m³ to 10⁻² m³ at 300 K against a constant pressure of 10⁵ Nm⁻². The work done on the gas is

Answer:

option (3) −900 J

Explanation:

The work done due to change in volume against a constant pressure is given by

⇒ W = −PΔV

Given,

P = 10⁵ Nm⁻²

$\sf V_i$ = 10⁻³ m³

$\sf V_f$ = 10⁻² m³

Substitute the values,

$\implies \sf W= -P(V_f -V_i)$

W = −10⁵ × (10⁻² − 10⁻³)

W = −10⁵ × (0.01 − 0.001)

W = −10⁵ × (0.009)

W = −10⁵ × 9 × 10⁻³

W = −9 × 10²

W = −900 J

The work done on the gas is −900 J

Answer 2

*Question →

An ideal gas expands isothermally from

10-3 m3 to 10-2 m3 at 300 K against a constant

pressure of 105 Nm-2. The work done on the gas is

• (1) - 900 kJ
• (2) + 270 kJ
• (3) - 900 J
• (4) + 900 kJ

*Answer →

Given that,

$P = 1 × 10⁵ N/m²$

$\sf V_i = 1 \times 10^{ - 3} {m}^{3}$

$\sf V_f = 1 \times 10^{ - 2} {m}^{3}$

$∆V = \sf V_f - \sf V_i = {10}^{ - 2} - 10^{ - 3} \\ = 9 \times {10}^{ - 3 } {m}^{3} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

As we know that,

• → W = - P ∆ V
• → W = - 1 × 10⁵ (10^-2 - 10^-3)
• → W = -10⁵ × (9 × 10^-3)
• → W = -9 × 10² = -900J

Hence the work done is -900J.

So option (3) -900J is correct.