Question

A 150 W electric heater operates for 12.0 min to heat an ideal gas in a cylinder. During this time, the gas expands from 3.00 L to 11.0 L against a constant pressure of 1.03 atm. What is the change in internal energy (in kJ) of the gas?

Answer 1

Answer: The change in internal energy of the gas is 108.835 kJ

Explanation:

To calculate the work done for reversible expansion process, we use the equation:

$W=P\Delta V=-P(V_2-V_1)$

where,

W = work done

P = pressure = 1.03 atm

$V_1$ = initial volume = 3.00 L

$V_2$ = final volume = 11.0 L

Putting values in above equation, we get:

$W=-(1.03)\times (11.0-3.00)=8.24L.atm=834.9J=0.835kJ$     (Conversion factor:  1 L. atm = 101.325 J)

Calculating the heat from power:

$Q=P\times t$

where,

Q = heat required

P = power = 150 W

t =  time = 12 min = 720 s       (Conversion factor:  1 min = 60 s)

Putting values in above equation:

$Q=150\times 720=108000J=108kJ$

The equation for first law of thermodynamics follows:

$Q=dU+W$

where,

Q = total amount of heat required = 108 kJ

dU = Change in internal energy = ?

W = work done  = -0.835 kJ

Putting values in above equation, we get:

$108kJ=dU+(-0.835)\\\\dU=(108+0.835)=108.835kJ$

Hence, the change in internal energy of the gas is 108.835 kJ