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

Step-by-step explanation:

The given data is as follows.

P = 3 atm

=
`3 atm * (1.01325 * 10^(5) Pa)/(1 atm)`

=
`3.03975 * 10^(5) Pa`

`V_(1)` = 9 L =
`9 * 10^(-3) m^(3)` (as 1 L = 0.001
`m^(3)`),

`V_(2)` = 15 L =
`15 * 10^(-3) m^(3)`

Heat energy = 800 J

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

W =
`P * \Delta V`

or, W =
`P * (V_(2) - V_(1))`

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

W =
`P * (V_(2) - V_(1))`

=
`3.03975 * 10^(5) Pa * (15 * 10^(-3) m^(3) - 9 * 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.