A cylinder with a moveable piston on top, free to move up and down, contains one mole of an ideal gas initially at a temperature of Ti = 4.8°C. The cylinder is heated at a constant pressure of 1.00 atm, and it expands to eight times its original volume.(a)Calculate the new temperature Tf of the gas (in K). K(b)Calculate the work done (in kJ) on the gas during the expansion. kJ
Question
A cylinder with a moveable piston on top, free to move up and down, contains one mole of an ideal gas initially at a temperature of Ti = 4.8°C. The cylinder is heated at a constant pressure of 1.00 atm, and it expands to eight times its original volume.(a)Calculate the new temperature Tf of the gas (in K). K(b)Calculate the work done (in kJ) on the gas during the expansion. kJ
Solution 1
(a) To calculate the new temperature Tf of the gas, we can use the ideal gas law, which states that the ratio of the volume and temperature of a gas is constant if the pressure and amount of gas are constant. This can be written as:
V1/T1 = V2/T2
where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature. We know that the volume increases to eight times its original volume, so V2 = 8V1. We also know that the initial temperature Ti = 4.8°C = 277.95 K (since the Kelvin scale starts at absolute zero, -273.15°C). Substituting these values into the equation gives:
1/277.95 = 8/T2
Solving for T2 gives T2 = 8 * 277.95 = 2223.6 K. So the final temperature of the gas is 2223.6 K.
(b) The work done on the gas during the expansion can be calculated using the formula for work done in an isobaric (constant pressure) process:
W = P * ΔV
where P is the pressure and ΔV is the change in volume. We know that the pressure P = 1.00 atm = 101.325 kPa (since 1 atm = 101.325 kPa), and the volume changes by a factor of 8, so ΔV = 8V1 - V1 = 7V1. However, we don't know the actual volume V1, so we can't calculate the work done in kJ. We would need more information to calculate this.
Solution 2
(a) To calculate the new temperature Tf of the gas, we can use the ideal gas law, which states that the ratio of the volume and temperature of a gas is constant if the pressure and amount of gas are constant. This can be written as:
V1/T1 = V2/T2
where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature. We know that the initial temperature Ti = 4.8°C = 277.95 K (since 0°C = 273.15 K), and the final volume is eight times the initial volume. Therefore, we can write:
1/277.95 = 8/T2
Solving for T2 gives:
T2 = 8 * 277.95 = 2223.6 K
So, the new temperature Tf of the gas is 2223.6 K.
(b) To calculate the work done on the gas during the expansion, we can use the formula for work done in a constant pressure process, which is:
W = P * ΔV
where P is the pressure and ΔV is the change in volume. We know that the pressure is 1.00 atm = 101.325 J/(m^3K) (since 1 atm = 101.325 J/(m^3K)), and the change in volume is V2 - V1 = 8V1 - V1 = 7V1. Therefore, we can write:
W = 101.325 * 7V1
However, we don't know the value of V1. But we know that the initial state of the gas follows the ideal gas law, which can be written as:
P * V1 = n * R * T1
where n is the number of moles of gas, R is the ideal gas constant, and T1 is the initial temperature. We know that n = 1 mole, R = 0.0821 Latm/(Kmol) (the value of the ideal gas constant in these units), and T1 = 277.95 K. Therefore, we can write:
V1 = n * R * T1 / P = 1 * 0.0821 * 277.95 / 1 = 22.8 L
Substituting this into the equation for work gives:
W = 101.325 * 7 * 22.8 = 16140.21 J = 16.14 kJ
So, the work done on the gas during the expansion is 16.14 kJ.
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