The work done during the expansion of a gas from a volume of $4$ dm $^{3}$ to $6$ dm $^{3}$ against a constant external pressure of $3$ atm is:

608 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

+ 304 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

304 J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 608 J

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $- 608 J$
Work done during expansion of a gas is:-
$W = - P △ V$
$W = - P (V_{f i n a l} - V_{i n i t i a l})$
$W = - 3 (6 - 4)$
$W = - 6 L - a t m = - 6 \times 101.32 =\approx - 608 J$ $(∵ 1 d m^{3} = 1 L)$

Suggest Corrections

Similar questions

4 d m^{3}

6 d m^{3}

3 a t m

4 {dm}^{3}

6 {dm}^{3}

3 atm

(1 L atm = 101.32 J)

4 {dm}^{3} to 6 {dm}^{3}

3 atm

1 L atm = 101.32 J

4 {d m}^{3}

6 {d m}^{3}

3 a t m

[1 L a t m = 101.32 J]

4 d m^{3} t o 6 d m^{3}

Join BYJU'S Learning Program