A gas is supplied with 1500 ; J of heat energ | vedclass

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(C) According to the first law of thermodynamics:$\Delta Q = \Delta U + \Delta W$Since the process occurs at constant pressure,the work done is $\Delt

Vedclass · Accepted Answer

$975$

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(C) According to the first law of thermodynamics:$\Delta Q = \Delta U + \Delta W$Since the process occurs at constant pressure,the work done is $\Delta W = P \Delta V$.Therefore,$\Delta Q = \Delta U + P \Delta V$.Given:$\Delta Q = 1500 \; J$$P = 2.1 	imes 10^{5} \; N/m^{2}$$\Delta V = 2.5 	imes 10^{-3} \; m^{3}$Calculating the work done:$\Delta W = P \Delta V = (2.1 	imes 10^{5}) 	imes (2.5 	imes 10^{-3}) = 525 \; J$.Now,calculating the change in internal energy:$\Delta U = \Delta Q - \Delta W = 1500 - 525 = 975 \; J$.

$A$ gas is supplied with $1500 \; J$ of heat energy at constant pressure. If the constant pressure of the gas is $2.1 \times 10^{5} \; N/m^{2}$ and the increase in volume is $2.5 \times 10^{-3} \; m^{3}$,then the increase in the internal energy of the gas is ...... $J$.

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