An ideal gaseous reaction occurs at a constant pressure of 35.0 atm and releases 55.8 kJ of heat. Before the reaction, the volume of the system was 8.20 L. After the reaction, the volume of the system was 3.01 L. Calculate the total change in internal energy for the system.
From first law of thermodynamics , we know that
dU = dq + dW
Here, dU is the change in internal Energy
dq is the amount of Heat exchange
dW is the amount of Work done
According to the IUPAC Conversation , Heat released by the system is regarded as the negative quantity .
Hence ,
dq= -55.8kJ = -55800 J
Work done against constant pressure .
We know that
dW= - P dV
dW =-P ( V_{2} - V_{1} )
Here , P = 35.0 atm , V_{2} = 3.01 L & V_{1} = 8.20 L
dW = - 35.0 atm ( 3.01- 8.20) L
dW = 181.65 L atm
1 L atm = 101.325 J
Hence we get
dW = 181.65 × 101.325 J
dW = 18405.686 J
Hence internal energy
dU = dq + dW
= -55800 J + 18405.686 J
= - 37394.31 J
= -37.39431 KJ
Hence total Change in internal energy = -37.39431 KJ
An ideal gaseous reaction occurs at a constant pressure of 35.0 atm and releases 55.8 kJ...
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