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(D) From the first law of thermodynamics,$\Delta Q = \Delta U + W$. Since no heat is supplied or extracted,$\Delta Q = 0$,which implies $\Delta U = -W

Vedclass · Accepted Answer

$4.5$

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(D) From the first law of thermodynamics,$\Delta Q = \Delta U + W$. Since no heat is supplied or extracted,$\Delta Q = 0$,which implies $\Delta U = -W$.The work done $W$ is the area under the $P-V$ graph. The area is a trapezoid with parallel sides $P_A = 10 	imes 10^3 \, Pa$ and $P_B = 50 	imes 10^3 \, Pa$,and height $\Delta V = (200 - 50) \, cc = 150 	imes 10^{-6} \, m^3$.$W = \frac{1}{2} 	imes (P_A + P_B) 	imes (V_A - V_B)$$W = \frac{1}{2} 	imes (10 + 50) 	imes 10^3 	imes (200 - 50) 	imes 10^{-6}$$W = \frac{1}{2} 	imes 60 	imes 10^3 	imes 150 	imes 10^{-6} = 30 	imes 150 	imes 10^{-3} = 4.5 \, J$.Since the process is from $A$ to $B$ (compression),the work done by the gas is negative $(W = -4.5 \, J)$.Therefore,$\Delta U = -W = -(-4.5 \, J) = 4.5 \, J$.

The pressure of a gas changes linearly with volume from $A$ to $B$ as shown in the figure. If no heat is supplied to or extracted from the gas,then the change in the internal energy of the gas will be $............\,J$.

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