2 moles of an ideal gas at 20k changes from a pressure of 15 atm to 3 atm at constant temperature. Calculate the work done by the gas.
At constant temperature, P*V = constant and
P dV = - V dP = -nRT/P*dP
Work done is therefore
Integral P dV = -(nRT) dP/P
= (-nRT) ln P2/P1
= nRT*ln(P1/P2)
= 2.0 moles*RT*ln5
To calculate the work done by the gas, we can use the equation:
Work = -P × ΔV
Where:
- Work is the work done by the gas
- P is the pressure
- ΔV is the change in volume
In this case, we are given:
- Initial pressure (P1) = 15 atm
- Final pressure (P2) = 3 atm
- Number of moles (n) = 2
- Temperature (T) is constant
First, let's calculate the change in volume (ΔV).
The ideal gas law equation can be rearranged to find the volume (V):
PV = nRT
Where:
- V is the volume
- R is the ideal gas constant (approximately 0.0821 L·atm/(mol·K))
Since the temperature is constant, we can write:
P1V1 = nRT (equation 1)
P2V2 = nRT (equation 2)
Divide equation 2 by equation 1:
(P2V2) / (P1V1) = (nRT) / (nRT)
Cancel out the variables and rearrange:
V2 / V1 = P2 / P1
Now, let's substitute the given values:
V2 / V1 = (3 atm) / (15 atm) = 1/5
Rearranging the equation:
V2 = (1/5) × V1
Since the number of moles (n) and temperature (T) remain constant, we can say:
ΔV = V2 - V1 = (1/5) × V1 - V1 = (-4/5) × V1
Now, let's calculate the work done by the gas using the equation:
Work = -P × ΔV
Work = -(P2 - P1) × ΔV
Substitute the given values:
Work = -(3 atm - 15 atm) × (-4/5) × V1
Work = 12 × (4/5) × V1
Work = 9.6 × V1
Therefore, the work done by the gas is 9.6 times the initial volume (V1) of the gas.