The internal energy of an ideal gas is given | vedclass

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(B) The internal energy of the gas is $U = 1.5 PV$. The change in internal energy is $\Delta U = 1.5 \Delta(PV) = 1.5 P \Delta V$ (since pressure $P$

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$5$

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(B) The internal energy of the gas is $U = 1.5 PV$. The change in internal energy is $\Delta U = 1.5 \Delta(PV) = 1.5 P \Delta V$ (since pressure $P$ is constant).Given $P = 2 	imes 10^5 \ Pa$,$V_i = 10 \ cm^3 = 10 	imes 10^{-6} \ m^3$,and $V_f = 20 \ cm^3 = 20 	imes 10^{-6} \ m^3$.Change in volume $\Delta V = V_f - V_i = 10 	imes 10^{-6} \ m^3$.Change in internal energy $\Delta U = 1.5 	imes (2 	imes 10^5) 	imes (10 	imes 10^{-6}) = 1.5 	imes 2 = 3 \ J$.Work done by the gas $W = P \Delta V = (2 	imes 10^5) 	imes (10 	imes 10^{-6}) = 2 \ J$.According to the First Law of Thermodynamics,$Q = \Delta U + W$.$Q = 3 \ J + 2 \ J = 5 \ J$.

The internal energy of an ideal gas is given by $U = 1.5 PV$. It expands from $10 \ cm^3$ to $20 \ cm^3$ against a constant pressure of $2 \times 10^5 \ Pa$. Heat absorbed by the gas in the process is (in $J$)

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