Chemistry, asked by LoverLoser, 4 hours ago

*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

Answers

Answered by snehitha2
21

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

Answered by Vikramjeeth
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.

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