A tank with a constant volume of 3.44 m3 contains 14 moles of a monatomic ideal gas. The gas is initially at a temperature of 300 K. An electric heater is used to transfer 52600 J of energy into the gas. It may help you to recall that CVCV = 12.47 J/K/mole for a monatomic ideal gas, and that the number of gas molecules is equal to Avagadros number (6.022 × 1023) times the number of moles of the gas.

1) What is the temperature of the gas after the energy is added?

2) What is the change in pressure of the gas?
3) How much work was done by the gas during this process?

1) Q = nc*changeT

changeT + 300 = Tf
2)P = nRT/V. Do for intial and final temps
3) None. No change in volume means no work

To find the temperature of the gas after the energy is added, you can use the equation for the change in internal energy:

ΔU = nCvΔT

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

First, let's calculate the change in internal energy. Since the gas is monatomic, Cv is equal to 12.47 J/K/mole. We are given that the number of moles of gas is 14, so:

ΔU = 14 * 12.47 J/K/mole * ΔT

Next, we know that the electric heater transferred 52600 J of energy into the gas:

ΔU = 52600 J

Now we can solve for ΔT:

ΔT = ΔU / (14 * 12.47 J/K/mole)

1) Calculate the temperature of the gas after the energy is added:

ΔT = 52600 J / (14 * 12.47 J/K/mole)
ΔT ≈ 266.08 K

The temperature of the gas after the energy is added is approximately 266.08 K.

2) To find the change in pressure 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 of gas, R is the ideal gas constant, and T is the temperature.

Since the volume of the tank is constant, the initial and final volumes are the same. We can rewrite the ideal gas law as:

P1 * V = P2 * V

where P1 is the initial pressure and P2 is the final pressure.

Since the number of moles and volume are constant, we can simplify the equation to:

P1 = P2

Therefore, the change in pressure is zero. The pressure of the gas remains constant.

3) The work done by the gas during this process can be calculated using the equation:

W = -ΔU = -PΔV

Since the volume is constant, ΔV is zero. Therefore, the work done by the gas is also zero. No work is done during this process.

To solve this problem, we can apply the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

1) To find the temperature of the gas after the energy is added, we can use the equation:

ΔU = Q - W

where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.

Since the volume is constant, there is no work done, so W = 0.

ΔU = Q - 0
ΔU = Q = 52600 J

The change in internal energy for a monatomic ideal gas is given by the equation:

ΔU = CVCV * n * ΔT

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

We can rearrange this equation to solve for ΔT:

ΔT = ΔU / (CVCV * n)

ΔT = 52600 J / (12.47 J/K/mole * 14 moles)

Calculating this expression gives us:

ΔT ≈ 300.57 K

Therefore, the temperature of the gas after the energy is added is approximately 300.57 K.

2) The change in pressure of the gas can be calculated using the ideal gas law:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.

Since the volume is constant, we can rewrite the equation as:

P1 = nRT1 / V

where P1 is the initial pressure and T1 is the initial temperature.

Similarly, the final pressure P2 can be given as:

P2 = nRT2 / V

where T2 is the final temperature.

To find the change in pressure, ΔP, we subtract the initial pressure from the final pressure:

ΔP = P2 - P1

ΔP = (nRT2 / V) - (nRT1 / V)

Plugging in the values:

ΔP = (14 moles * 8.314 J/(mol*K) * 300.57 K) / 3.44 m^3
ΔP ≈ 268.99 Pa

Therefore, the change in pressure of the gas is approximately 268.99 Pa.

3) The work done by the gas during this process can be calculated using the equation:

W = -PΔV

where W is the work done, P is the pressure, and ΔV is the change in volume.

Since the volume is constant, ΔV = 0, so the work done is:

W = -P * 0
W = 0

Therefore, the work done by the gas during this process is zero.