Thermodynamics Question 6
Question 6 - 25 January - Shift 2
Match List I with List II :
| List I | List II | ||
|---|---|---|---|
| A. | Isothermal math Process |
I. | Work done by the gas decreases internal energy |
| B. | Adiabatic Process | II. | No change in internal energy |
| C. | Isochoric Process | III. | The heat absorbed goes partly to increase internal energy and partly to do work |
| D. | Isobaric Process | IV. | No work is done on or by the gas |
Choose the correct answer from the options given below:
(1) A-II, B-I, C-III, D-IV
(2) A-II, B-I, C-IV, D-III
(3) A-I, B-II, C-IV, D-III
(4) A-I, B-II, C-III, D-IV
Show Answer
Answer: (2)
Solution:
Formula: Isothermal process, Isobaric process, Isochoric process, Adiabatic process
$\Delta U=nC_v \Delta T$
For isothermal process $T$ is constant
So $\Delta U=0$
$A \longrightarrow II$
Adiabatic process
$\Delta Q=0$
$\Delta Q=\Delta U+\Delta W$
$ \Delta U=-\Delta W $
Work done by gas is positive
So $\Delta U$ is negative
$B \longrightarrow I$
For Isochoric process $\Delta W=0$
$C \longrightarrow IV$
For Isobaric process
$\Delta W=P \Delta V \neq 0$
$\Delta U=nC_V \Delta T \neq 0$
Heat absorbed goes partly to increase internal energy and partly do work.
