Question

Under isothermal condition a gas at 300k expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is

Answer 1

workdone by the gas in isothermal process is given by, $W=-P_{ext}(V_2-V_1)$

[ it is irreversible isothermal process. ]

here, $P_{ext}$ = 2bar = 2 × 10^5 Pascal or 2 × 10^5 N/m²

[ as you know, 1 bar = 10^5 Pascal ]

$V_2$ = 0.25L = 0.25 × 10^-3 m³

$V_1$ = 0.1L = 0.1 × 10^-3 m³

[ as you also know, 1000L = 1m³ ]

so, workdone by the gas , W = -2 × 10^5(0.25 - 0.10) × 10^-3 Nm or Joule

= -2 × (0.15) × 10^2 joule

= -30 joule

hence answer will be -30J

Answer 2

Question: Under isothermal condition a gas at 300k expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is:

Solution: $-30J$

Step by step explanation:

$W_i_r_r= -P_e_x_t$ ∆V

$=-2 \: bar \times (0.25-0.1)L\\=-2 \times 0.15 \: Lbar\\=-0.30 \: Lbar\\=-0.30 \times 100J\\= -30 J$

Hence, the work done by gas is,

$\boxed{-30J}$