Question

An ideal gas at temperature T0 is slowly compressed at constant pressure of 2 atm from a volume of 10 liters to a volume of 2 liters. Then the volume of the gas is held constant while heat is added, raising the gas temperature back to T0. Calculate the work done ON the gas. 1 atm = 1.0×10^5 Pascals and 1 liter = 0.001 m^3.

1. +800 J
2. −400 J
3. +400 J
4. −800 J
5. −1600 J
6. +1600 J

Answer 1

To calculate the work done on the gas during the compression process at constant pressure, we can use the formula for work done in a constant pressure process:

\[
W = -P \Delta V
\]

where:
- \(W\) is the work done on the gas,
- \(P\) is the pressure,
- \(\Delta V\) is the change in volume.

First, convert the pressure from atm to Pascals:

\[
P = 2 \, \text{atm} \times 1.0 \times 10^5 \, \text{Pa/atm} = 2.0 \times 10^5 \, \text{Pa}
\]

Next, calculate the change in volume:

\[
\Delta V = V_f - V_i = 2 \, \text{liters} - 10 \, \text{liters} = 2 \, \text{liters} - 10 \, \text{liters} = -8 \, \text{liters}
\]

Convert the volume change to cubic meters:

\[
\Delta V = -8 \, \text{liters} \times 0.001 \, \text{m}^3/\text{liter} = -0.008 \, \text{m}^3
\]

Now, we can plug in the values into the work formula:

\[
W = -P \Delta V = - (2.0 \times 10^5 \, \text{Pa}) \times (-0.008 \, \text{m}^3)
\]

Calculating this gives:

\[
W = (2.0 \times 10^5) \times (0.008) = 1600 \, \text{J}
\]

Since this is the work done ON the gas, it is positive when considering work done on the gas (the work is done on it):

\[
W = +1600 \, \text{J}
\]

Thus, the work done on the gas is:

\[
\boxed{+1600 \, \text{J}}
\]