A perfect gas goes from state A to another state B by absorbing $8 \times 10^{5} \mathrm{J}$ of heat and doing $6.5 \times 10^{5} \mathrm{J}$ of external work. It is now transferred between the same two states in another process in which it absorbs $10^{5} J$ of heat. Then in the second process
Text Solution
Verified by ExpertsA
In first process using $\Delta Q = \Delta U + \Delta W$
⇒ ⇒ $8 \times 10^{5} = \Delta U + 6.5 \times 10^{5}$ ⇒ ⇒ $\Delta U = 1.5 \times 10J$
Since final and initial states are same in both process
So $\Delta U$ will be same in both process
For second process using $\Delta Q = \Delta U + \Delta W$
⇒ ⇒ $10^{5} = 1.5 \times 10^{5} + \Delta W$ ⇒ ⇒ $\Delta W = -0.5 \times 10^{5} J$
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems