An ideal gas takes part in two thermal process in which it is heated from same initial state A. Once it is taken from A to B and in the other case from A to C. The points B and C lie on an isotherm. Choose the correct options.

Options

A. Change in internal energy for both the process is the same.
B. the heat supplied to gas in `A to B` is greater than that in `A to C`.
C. The heat supplied to gas in `A to B` is less than that is `A to C`.
D. The work done by the gas in `A to B` is equal to than that in `A to C`

Answer

A, C

Explanation

`W=`Area of trapezium
`=(1)/(2)(P_(0)+(C)/(V))(V-V_(0))`
`(dW)/(dV)=(1)/(2)(P_(0)+(C)/(V)+(1)/(2)(V-V_(0))(-(C)/(V^(2)))`
`(1)/(2)(P_(0)+(C)/(V)-(C)/(V_(2))(V-V_(0)))`
`=(1)/(2)(P_(0)+(C)/(V)-(C)/(V)+(CV_(0))/(V^(2)))=(P_(0)+(CV_(0))/(V^(2)))`
Since `(dW)/(dV)` is always positive,
Now, work done in process AC is greater than that of AB.
Hence heat supplied in process AC will be more than that of AB.