1. Calculate the work done on the system when 1.00 mol of gas held behind a piston expands irreversibly from volume of 1.00dm cube to a volume 10.00dm cube against an external pressure of 1.00 bar.

2. A perfect gas expand reversibly at a constant temperature of 298k so that its volume doubles. What is the change in the molar internal energy of the gas?

1. Calculate the work done on the system when

1.00 mol of gas held behind a piston expands
irreversibly from volume of 1.00dm cube to a
volume 10.00dm cube against an external
pressure of 1.00 bar.
2. A perfect gas expand reversibly at a constant
temperature of 298k so that its volume doubles.
What is the change in the molar internal energy of
the gas?

1. To calculate the work done on the system during an irreversible expansion, we can use the formula:

Work = -Pext * ΔV

where Pext is the external pressure and ΔV is the change in volume. In this case, Pext = 1.00 bar and ΔV = (10.00 dm^3) - (1.00 dm^3) = 9.00 dm^3.

However, we need to convert the units so that they are consistent. 1 bar is equal to 100,000 Pa, and 1 dm^3 is equal to 1 liter, which is equal to 0.001 m^3. Therefore, the external pressure is 1.00 bar * (100,000 Pa/bar) = 100,000 Pa, and the change in volume is 9.00 dm^3 * (0.001 m^3/dm^3) = 0.009 m^3.

Now, we can calculate the work done:

Work = -Pext * ΔV
= -(100,000 Pa) * (0.009 m^3)
= -900 J

So, the work done on the system during the irreversible expansion is -900 J.

2. For a perfect gas that undergoes a reversible expansion at constant temperature, the change in molar internal energy (ΔU) is given by:

ΔU = nCvΔT

where n is the number of moles of gas, Cv is the molar specific heat capacity at constant volume, and ΔT is the temperature change.

Since the gas is expanding reversibly at constant temperature, ΔU = 0, because there is no change in molar internal energy. This is because in a reversible process, the gas exchanges heat with its surroundings to maintain constant temperature, and there is no net energy transfer in the form of work done on or by the gas.

Therefore, the change in the molar internal energy of the gas is 0.

To calculate the work done on the system and the change in molar internal energy of a gas, we will need to use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (Q) minus the work done by the system (W):

ΔU = Q - W

Let's solve the two questions step by step:

1. To calculate the work done on the system when 1.00 mol of gas expands irreversibly from 1.00 dm^3 to 10.00 dm^3 against an external pressure of 1.00 bar:

First, we need to convert the pressure from bar to pascals (Pa):
1 bar = 100,000 Pa

The work done on the system (W) can be calculated using the formula:
W = -Pext * ΔV

Where Pext is the external pressure and ΔV is the change in volume. We need to be careful with signs, as work done on the system is considered positive.

Given:
Pext = 1.00 bar = 100,000 Pa
Initial volume (V1) = 1.00 dm^3 = 0.001 m^3
Final volume (V2) = 10.00 dm^3 = 0.01 m^3

ΔV = V2 - V1 = 0.01 m^3 - 0.001 m^3 = 0.009 m^3

Plugging in the values:
W = -Pext * ΔV = -(100,000 Pa) * (0.009 m^3) = -900 J

Therefore, the work done on the system when 1.00 mol of gas expands irreversibly is -900 J.

2. To calculate the change in molar internal energy of the gas when it expands reversibly at a constant temperature of 298 K and doubles its volume:

For an ideal gas, the change in internal energy (ΔU) only depends on temperature (ΔT), not on the path of the process. Since the process is reversible, no work is done on or by the system, so:
ΔU = Q

Given that the temperature (T) is constant:
ΔT = T2 - T1 = 298 K - 298 K = 0 K

Since ΔT = 0, there is no change in molar internal energy (ΔU = 0).

Therefore, the change in molar internal energy of the gas is 0.