A cylinder with a moveable piston holds 2.60 mol of argon at a constant temperature of 295 K. As the gas is compressed isothermally, its pressure increases from 101 kPa to 149 kPa.
(a) Find the final volume of the gas.
Answer in m3
(b) Find the work done by the gas.
Answer in kJ
(c) Find the heat added to the gas.
Answer in kJ
To solve this problem, we can use the ideal gas law and the formula for work done by a gas.
(a) To find the final volume of the gas, we can use the ideal gas law:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
We can rearrange the formula:
V = (nRT) / P
Given:
P1 = 101 kPa
P2 = 149 kPa
n = 2.60 mol
T = 295 K
R = 8.314 J/(mol·K)
First, convert the pressures to Pascal:
P1 = 101 kPa × 1000 Pa/kPa = 101,000 Pa
P2 = 149 kPa × 1000 Pa/kPa = 149,000 Pa
Plugging in the values:
V1 = (2.60 mol × 8.314 J/(mol·K) × 295 K) / 101,000 Pa
V2 = (2.60 mol × 8.314 J/(mol·K) × 295 K) / 149,000 Pa
Calculating V2:
V2 = (2.60 mol × 8.314 J/(mol·K) × 295 K) / 149,000 Pa
≈ 0.0317 m^3
Therefore, the final volume of the gas is approximately 0.0317 m^3.
(b) To find the work done by the gas, we can use the formula:
W = -PΔV
Where P is the average pressure and ΔV is the change in volume.
Given:
P1 = 101,000 Pa
P2 = 149,000 Pa
V1 = unknown
V2 = 0.0317 m^3
The average pressure is the average of the initial and final pressures:
Pavg = (P1 + P2) / 2
Plugging in the values:
Pavg = (101,000 Pa + 149,000 Pa) / 2
= 125,000 Pa
Calculating the work done:
W = -Pavg ΔV
= -(125,000 Pa)(0.0317 m^3)
≈ -3,962.5 J
≈ -3.96 kJ
Therefore, the work done by the gas is approximately -3.96 kJ.
(c) Since the process is isothermal, the temperature remains constant, so there is no change in the internal energy. Therefore, the heat added to the gas is equal to the work done:
Q = W = -3.96 kJ
Therefore, the heat added to the gas is approximately -3.96 kJ. Note that the negative sign indicates that heat is being removed from the gas.
To find the answers to these questions, we can use the ideal gas law and the formula for work done by a gas in an isothermal process.
(a) The ideal gas law equation is:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Since the temperature, number of moles, and gas constant remain constant, we can rewrite the equation as:
P1V1 = P2V2
Where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.
We are given P1 = 101 kPa, P2 = 149 kPa, and n = 2.60 mol.
Plugging in the values and solving for V2, we get:
101 kPa * V1 = 149 kPa * V2
V2 = (101 kPa * V1) / 149 kPa
Now we need to convert the pressure from kilopascals to pascals since the SI unit for pressure is pascal (Pa).
1 kPa = 1000 Pa
So, P1 = 101 kPa * 1000 Pa/kPa = 101000 Pa
And P2 = 149 kPa * 1000 Pa/kPa = 149000 Pa
Now we can substitute the values and calculate V2:
V2 = (101000 Pa * V1) / 149000 Pa
(b) The work done by the gas can be calculated using the formula:
W = -P∆V
Where W is the work done, P is the pressure, and ∆V is the change in volume.
In this case, the gas is compressed, so V1 > V2. Therefore, we can rewrite the formula as:
W = P∆V
Since the process is isothermal, the temperature remains constant, and we know that ∆V = V2 - V1, so:
W = P(V2 - V1)
Now we can substitute the given values and calculate the work done.
(c) The heat added to the gas can be calculated using the first law of thermodynamics:
∆Q = ∆U + W
Where ∆Q is the heat added, ∆U is the change in internal energy, and W is the work done.
In an isothermal process, the change in internal energy (∆U) is zero. Therefore:
∆Q = W
Now we can substitute the value of W that we calculated in part (b) to find the heat added to the gas.