Question

What is the change in internal energy when a gas contracts from 377mL to 177mL under a constant pressure of 1520 mm of Hg, while at the same time 124J heat is released?
A.40.52 J
B.\[ - 248\] J
C.\[ + 275.3\] J
D.\[ - 83.48\] J

Answer 1

Hint:Since, we know that the first law of thermodynamics the first law of thermodynamics states that the change in internal energy of a system \[\Delta U\], U equals the net heat transfer into the system Q, plus the net work done on the system W. In equation form, the first law of thermodynamics is, \[\Delta U = q + w\]

Complete step by step answer:
According to first law of thermodynamics,
\[\Delta U = q + w\]
Where,
\[\Delta U\] = Change in internal energy
q = heat absorbed or released
w = work done on or by the system
w = work done on the system \[ = - P\Delta V\] {Work is done on the system as the final volume is lesser than initial volume and is positive}
\[\Delta H = - 124\;J\] (Since, the heat is removed)
\[{V_1} = 377\] mL
\[{V_2}\; = 177\] mL
\[\therefore \Delta V = {V_2} - {V_1}\]
\[ = 177-377\]
\[ = - 200\] mL
\[ = - 0.2\;\]L
Pressure, \[P = 1520torr = \dfrac{{1520}}{{760}}\] \[ = 2\;atm\](Since, \[760\,torr = 1\,atm\])
\[P.\Delta V = 2 \times ( - 0.2) = - 0.4\;L\;atm\]
As we know that,
\[1\;L\;atm = 101.3\;J\]
\[\therefore P.\Delta V = - 0.4\;L\;atm\] \[ = ( - 0.4 \times 101.3)\;J\] \[ = - 40.52\;J\]
As we know that,
\[\Delta U = q-P\Delta V\]
Substituting the values in the above equation, we get,
\[\Delta U = - 124 - ( - 40.52)\]
\[ \Rightarrow \Delta U = - 124 + 40.52 = - 83.48\;J\]
Hence, the change in the internal energy will be −83.48 J.

Therefore, the correct answer is option (D).

Note:
The first law of thermodynamics applies the conservation of energy principle to systems where heat transfer and doing work are the methods of transferring energy into and out of the system.