Question

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.

Answer 1

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