First slide

First law of thermodynamics

Question

In a thermodynamic process the pressure of a fixed mass of a gas is changed in such a manner that the gas molecules give out 30 J of heat and 10 J of work done on the gas. If the initial internal energy of the gas was 40 J. Then the final internal energy will be ?

Moderate

A

80 J
B

20 J
C

Zero
D

-20 J

Solution

$\begin{array}{rcl} first law of thermodynamics is dQ=dU + dW \\ d Q i s h e a t e n e r g y & = & - 30 J (- v e \sin c e h e a t g i v e n o u t b y s y s t e m) \\ d W i s w o r k d o n e & = & - 10 J (- v e \sin c e w o r k i s d o n e o n s y s t e m) \\ d U i s c h a n g e i n i n t e r n a l e n e r g y & = & U_{f} - U_{i} i {=U}_{final} {-U}_{initial} here U_{i} = 40 J \\ s u b s t i t u t e t h e v a l u e s i n f i r s t l a w o f t h e r m o d y n a m i c s \\ dQ= [U_{f} {-U}_{i}] +dW \\ \Rightarrow & -30= [U_{f} - 40] - 10 \\ \Rightarrow & {-30+50=U}_{f} \\ \Rightarrow & U_{f} =20 J \end{array}$

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