Question

Ideal He gas expanded at constant pressure of 3 atm until its volume was increased from 9 liters to 15 liters. During this process, the gas absorbed 800J of heat from the surroundings. Please calculate the internal energy change of the gas, AE.

Answer 1

Explanation:

The given data is as follows.

             P = 3 atm

                = $3 atm \times \frac{1.01325 \times 10^{5} Pa}{1 atm}$  

                 = $3.03975 \times 10^{5} Pa$

    $V_{1}$ = 9 L = $9 \times 10^{-3} m^{3}$    (as 1 L = 0.001 $m^{3}$),  

        $V_{2}$ = 15 L = $15 \times 10^{-3} m^{3}$

            Heat energy = 800 J

As relation between work, pressure and change in volume is as follows.

                  W = $P \times \Delta V$

or,                W = $P \times (V_{2} - V_{1})$

Therefore, putting the given values into the above formula as follows.

                  W = $P \times (V_{2} - V_{1})$

                      = $3.03975 \times 10^{5} Pa \times (15 \times 10^{-3} m^{3} - 9 \times 10^{-3} m^{3})$

                      = 1823.85 Nm

or,                   = 1823.85 J

As internal energy of the gas $\Delta E$ is as follows.

                     $\Delta E$ = Q - W

                                  = 800 J - 1823.85 J

                                  = -1023.85 J

Thus, we can conclude that the internal energy change of the given gas is -1023.85 J.