A gas under constant pressure of 45 × 105 pa when subjected to 800 k j of heat, changes the volume from 0.5 m3 to 2.0 m3 . The change in internal energy of the gas is

Question

A gas under constant pressure of 45 × 105 pa when subjected to 800 k j of heat, changes the volume from 0.5 m3 to 2.0 m3 . The change in internal energy of the gas is (A)  6.75 × 105 J  (B)  5.25 × 105 J  (C)  3.25 × 105 J  (D)  1.25 × 105 J

Tardigrade Admin · Accepted Answer

Given, p = 4.5 × 102 Pa dQ =800 kJ = 800 × 103 J Change in volume = (2.0 - 0.5) m3 = 1.5 m3 From first law of thermodynamics, dQ = dU + p ⋅ dV dU = dQ - p ⋅ dV = 800 × 103 - 4 .5 × 105 × 1.5 = - 1.25 × 105 J Change in internal energy (V)= 1.25 × 105 J.

Q. A gas under constant pressure of $45 \times 1 0^{5} p_{a}$ when subjected to $800 kj$ of heat, changes the volume from $0.5 m^{3}$ to $2.0 m^{3}$ . The change in internal energy of the gas is

$6.75 \times 1 0^{5} J$

$5.25 \times 1 0^{5} J$

$3.25 \times 1 0^{5} J$

$1.25 \times 1 0^{5} J$

Q. A gas under constant pressure of 45×105pa​ when subjected to 800kj of heat, changes the volume from 0.5m3 to 2.0m3 . The change in internal energy of the gas is

6.75×105J

5.25×105J

3.25×105J

1.25×105J

Given, p=4.5×102Pa dQ=800kJ =800×103J Change in volume =(2.0−0.5)m3 =1.5m3 From first law of thermodynamics, dQ=dU+p⋅dV dU=dQ−p⋅dV =800×103−4.5×105×1.5 =−1.25×105J Change in internal energy (V)=1.25×105J.

Q. A gas under constant pressure of $45 \times 1 0^{5} p_{a}$ when subjected to $800 kj$ of heat, changes the volume from $0.5 m^{3}$ to $2.0 m^{3}$ . The change in internal energy of the gas is

$6.75 \times 1 0^{5} J$

$5.25 \times 1 0^{5} J$

$3.25 \times 1 0^{5} J$

$1.25 \times 1 0^{5} J$