Q51 / 180 · NEET 2019 (All India) · Chemistry

Under the isothermal condition, a gas at $300 K$ expands from $0.1 L$ to $0.25 L$ against a constant external pressure of $2$ bar. The work done by the gas is
(Given that $1 L$ bar $= 100 J$ )

Correct answer
A
$- 30 J$
B
$5 k J$
C
$25 J$
D
$30 J$

Solution

The work done by a gas,

W = P_{external} \times ΔV

= - 2 \times (v_{2} - v_{1})

= - 2 \times (0.25 - 0.10)

= - 0.30 L

bar

Given,

1 L

bar

= 100 J

.

So,

W = - 0 .3 \times 100 = - 30 J

.

Hence, the correct answer is option: 'A'.

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