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Physics Question on Thermodynamics

A gas under constant pressure of 45×105 pa45 \times 10^5\, p_a when subjected to 800 kj800 \,k j of heat, changes the volume from 0.5 m30.5\, m^3 to 2.0 m32.0\, m^3 . The change in internal energy of the gas is

A

6.75×105 J6.75 \times 10^5 \,J

B

5.25×105 J5.25 \times 10^5 \,J

C

3.25×105 J3.25 \times 10^5 \,J

D

1.25×105 J1.25 \times 10^5 \,J

Answer

1.25×105 J1.25 \times 10^5 \,J

Explanation

Solution

Given, p=4.5×102 Pap = 4.5 \times 10^2\, Pa
dQ=800 kJdQ =800 \,kJ
=800×103 J= 800 \times 10^3 \,J
Change in volume =(2.0−0.5)m3= (2.0 - 0.5) m^3
=1.5 m3= 1.5 \,m^3
From first law of thermodynamics,
dQ=dU+p⋅dVdQ = dU + p \cdot dV
dU=dQ−p⋅dVdU = dQ - p \cdot dV
=800×103−4.5×105×1.5= 800 \times 10^3 - 4 .5 \times 10^5 \times 1.5
=−1.25×105 J= - 1.25 \times 10^5\, J
Change in internal energy
(V)=1.25×105 J(V)= 1.25 \times 10^5\, J.