3.2 m 3 of a gas is heated at the pressure 105 N / m 2 until its volume increases by 25 %. Then, the external work done by the gas is × 104 J.

Question

3.2 m 3 of a gas is heated at the pressure 105 N / m 2 until its volume increases by 25 %. Then, the external work done by the gas is × 104 J. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

External work done = PdV = P ( V 2- V 1) V 1= V =3.2 m 3 and V 2 = V +25 % V = V +(25 V /100) = V +0.25 V ∴ W = P ( V +0.25 V - V ) =0.25 PV =0.25 × 105 × 3.2 =0.8 × 105 ∴ W =8 × 104 J

Q. $3.2 m^{3}$ of a gas is heated at the pressure $1 0^{5} N / m^{2}$ until its volume increases by $25%$ . Then, the external work done by the gas is ________ $\times 1 0^{4} J$ .

External work done $= P d V = P (V_{2} - V_{1})$
$V_{1} = V = 3.2 m^{3}$ and
$V_{2} = V + 25% V$
$= V + \frac{25 V}{100}$
$= V + 0.25 V$
$∴ W = P (V + 0.25 V - V)$
$= 0.25 P V$
$= 0.25 \times 1 0^{5} \times 3.2$
$= 0.8 \times 1 0^{5}$
$∴ W = 8 \times 1 0^{4} J$

Q. 3.2m3 of a gas is heated at the pressure 105N/m2 until its volume increases by 25%. Then, the external work done by the gas is ________×104J.

External work done =PdV=P(V2​−V1​) V1​=V=3.2m3 and V2​=V+25%V =V+10025V​ =V+0.25V ∴W=P(V+0.25V−V) =0.25PV =0.25×105×3.2 =0.8×105 ∴W=8×104J

Q. $3.2 m^{3}$ of a gas is heated at the pressure $1 0^{5} N / m^{2}$ until its volume increases by $25%$ . Then, the external work done by the gas is ________ $\times 1 0^{4} J$ .

External work done $= P d V = P (V_{2} - V_{1})$
$V_{1} = V = 3.2 m^{3}$ and
$V_{2} = V + 25% V$
$= V + \frac{25 V}{100}$
$= V + 0.25 V$
$∴ W = P (V + 0.25 V - V)$
$= 0.25 P V$
$= 0.25 \times 1 0^{5} \times 3.2$
$= 0.8 \times 1 0^{5}$
$∴ W = 8 \times 1 0^{4} J$