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(A) Since the container is insulated,the process is adiabatic,so $\delta Q = 0$. According to the first law of thermodynamics,$\Delta U = \delta Q + \

Vedclass · Accepted Answer

$-625$

Vedclass · Answer

(A) Since the container is insulated,the process is adiabatic,so $\delta Q = 0$. According to the first law of thermodynamics,$\Delta U = \delta Q + \delta W$. Here,$\delta W = -P_{ext} 	imes \Delta V$. Given: $P_{ext} = 2.5 \ bar = 2.5 	imes 10^5 \ Pa$,$V_1 = 4.5 \ dm^3 = 4.5 	imes 10^{-3} \ m^3$,$V_2 = 7 	imes 10^{-3} \ m^3$. $\Delta V = (7 - 4.5) 	imes 10^{-3} \ m^3 = 2.5 	imes 10^{-3} \ m^3$. $\Delta U = 0 + [-(2.5 	imes 10^5 \ Pa) 	imes (2.5 	imes 10^{-3} \ m^3)]$. $\Delta U = -6.25 	imes 10^2 \ J = -625 \ J$.

$A$ gas is allowed to expand in an insulated container against a constant external pressure of $2.5 \ bar$ from $4.5 \ dm^3$ to $7 \times 10^{-3} \ m^3$. What is the change in internal energy of the gas (in $J$)?

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