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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 = 8.0°C. The cylinder is heated at a constant pressure of 1.00 atm, and it expands to five 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. How is the work related to pressure and volume change? From the ideal gas law, how is the volume related to temperature, pressure, and number of moles? Can you use the ideal gas law to write the work expression in terms of known quantities? Be careful with signs and with unit conversions. 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 = 8.0°C. The cylinder is heated at a constant pressure of 1.00 atm, and it expands to five 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. How is the work related to pressure and volume change? From the ideal gas law, how is the volume related to temperature, pressure, and number of moles? Can you use the ideal gas law to write the work expression in terms of known quantities? Be careful with signs and with unit conversions. kJ

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Solution

(a) The ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

First, we need to convert the initial temperature from Celsius to Kelvin. The conversion is T(K) = T(°C) + 273.15, so Ti = 8.0°C + 273.15 = 281.15 K.

Since the pressure and the number of moles are constant, we can write the ideal gas law for the initial and final states as:

V1/T1 = V2/T2

We know that V2 = 5V1, so we can substitute this into the equation:

V1/T1 = 5V1/T2

Solving for T2 gives us:

T2 = 5T1 = 5 * 281.15 K = 1405.75 K.

(b) The work done on the gas during the expansion at constant pressure is given by W = PΔV, where ΔV is the change in volume.

We know that the final volume is five times the initial volume, so ΔV = V2 - V1 = 5V1 - V1 = 4V1.

The pressure is given in atm, so we need to convert it to the SI unit of pressure, which is pascals (Pa). 1 atm = 101325 Pa.

So, W = PΔV = 101325 Pa * 4V1.

However, we don't know the value of V1. But we can express it in terms of the other known quantities using the ideal gas law:

V1 = nRT1/P = (1 mol)(8.3145 J/(mol*K))(281.15 K)/(101325 Pa) = 0.023 m^3.

Substituting this into the equation for W gives us:

W = 101325 Pa * 4 * 0.023 m^3 = 9314.7 J.

Since 1 kJ = 1000 J, we can convert this to kJ to get:

W = 9.3147 kJ.

The work done on the gas during the expansion is 9.3147 kJ.

This problem has been solved

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