Air is expanded from 50 L to 150 L at 2 atm pressure. The external work done is (1 atm =1 × 105 N / m 2)

Question

Air is expanded from 50 L to 150 L at 2 atm pressure. The external work done is (1 atm =1 × 105 N / m 2) (A) 2 × 10-8 J (B) 2 × 104 J (C) 200 J (D) 2000 J

Tardigrade Admin · Accepted Answer

Work done W= pressure (p) × volume-change (Δ V) Given, p=2 atm =2 × 105 N / m 2 Δ V =V2-V1 =150-50=100 L =100 × 10-3 m 3 ∴ W =p Δ V ∴ W =2 × 105 × 100 × 10-3 =2 × 104 J

Q. Air is expanded from $50 L$ to $150 L$ at $2 a t m$ pressure. The external work done is $(1 a t m = 1 \times 1 0^{5} N / m^{2})$

$2 \times 1 0^{- 8} J$

$2 \times 1 0^{4} J$

$200 J$

$2000 J$

Work done
$W =$ pressure $(p) \times$ volume-change $(Δ V)$
Given, $p = 2 a t m = 2 \times 1 0^{5} N / m^{2}$
$Δ V = V_{2} - V_{1}$
$= 150 - 50 = 100 L$
$= 100 \times 1 0^{- 3} m^{3}$
$∴ W = p Δ V$
$∴ W = 2 \times 1 0^{5} \times 100 \times 1 0^{- 3}$
$= 2 \times 1 0^{4} J$

Q. Air is expanded from 50L to 150L at 2atm pressure. The external work done is (1atm=1×105N/m2)

2×10−8J

2×104J

200J

2000J