A gas takes part in two processes in which it is heated from the same initial state 1 to the same final temperature. The processes are shown on the p-V diagram by the straight lines 1-3 and 1-2. 2 and 3 are the points on the same isothermal curve. `Q_1` and `Q_2` are the heat transfer along the two processes. Then,

Options

A. `Q_1=Q_2`
B. `Q_1ltQ_2`
C. `Q_1gtQ_2`
D. Insufficient data

Answer

C

Explanation

`T_2=T_3` (lying on same isotherm)
`:.` `DeltaU_1=DeltaU_2`
W=Hatched area
`=1/2(p_i+p_f)(V_f-V_i)`
`=1/2(p_iV_f-p_iV_i+p_fV_f-p_fV_i)`

Now, `p_iV_i` and `p_fV_f` are same for both processes. Further `p_i` and `V_i`, area also same for both processes.
For `1rarr3 V_f` is more and `p_f` is less. Hence, W will be more.
Now, `Q=W+DeltaU`
`W_13gtW_12`
`:.` `Q_13gtQ_12` (as `DeltaU` is same)