Question

A 100 W electric heater (1 W = 1 J/s) operates for 11 min to heat the gas in a cylinder. At the same time, the gas expands from 1 L to 6 L against a constant external pressure of 3.527 atm. What is the change in internal energy of the gas? (1 L·atm = 0.1013 kJ)

Answer 1

Answer:

$\Delta U = 64218.9 J$

Explanation:

As we know that power of the heater is given as

P = 100 W

now the energy given by the heater for 11 min of time

$E = P \times t$

$E = 100 \times 11 \times 60$

$E = 100\times 11 \times 60$

$E = 66000 J$

now from 1st law of thermodynamics we know that

$E = \Delta U + W$

work done under constant pressure condition we have

$W = P \Delta V$

$W = (3.527 \times 1.01 \times 10^5)(6 - 1) \times 10^{-3}$

$W = 1781.13 J$

now from first equation we have

$66000 = 1781.13 + \Delta U$

$\Delta U = 64218.9 J$