I need urgent
Dear student,
Work done = change in KE
Work done by an ideal gas = P ∆V = nR ∆T
We are given: n = 10 mole
R = 1.9872 Cal/ mole.K
Let initial temperature be x°C ( or x+273 K). Final temperature is x+10°C ( or x+ 274 K). Hence ∆T = 10°C or 10 K
Increase in KE = 10* 1.9872 * 10 = 198.72 Cal
Approximately the correct answer is option (1) 200 Cal.
Regards
Work done = change in KE
Work done by an ideal gas = P ∆V = nR ∆T
We are given: n = 10 mole
R = 1.9872 Cal/ mole.K
Let initial temperature be x°C ( or x+273 K). Final temperature is x+10°C ( or x+ 274 K). Hence ∆T = 10°C or 10 K
Increase in KE = 10* 1.9872 * 10 = 198.72 Cal
Approximately the correct answer is option (1) 200 Cal.
Regards