A rigid container of nitrogen gas (N2) at 24° C contains 440 L at a pressure of 3.5 atm. If 26.6 kJ of heat are added to the container, what will be the new temperature of the gas?
° C
To determine the new temperature of the gas, we can use the ideal gas law equation:
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 in Kelvin.
First, let's convert the initial temperature from Celsius to Kelvin:
T1 = 24°C + 273.15 = 297.15 K
We can calculate the initial number of moles, n1, using the ideal gas law equation:
n1 = PV1 / (RT1)
Given:
P1 = 3.5 atm
V1 = 440 L
R = 0.0821 L.atm/(K.mol)
n1 = (3.5 atm * 440 L) / (0.0821 L.atm/(K.mol) * 297.15 K) = 54.69 mol
Now, let's determine the final number of moles, n2. Since the moles remain constant in this problem, n2 = n1 = 54.69 mol.
Next, we can express the final temperature, T2, using the ideal gas law equation:
T2 = (PV2) / (n2R)
We need to calculate the final pressure, V2, using the following formula:
V2 = (PV1 * T2) / (P1 * T1)
We know that the heat added to the container can be written as the formula:
Q = nCΔT
where Q is the heat, n is the number of moles, C is the molar heat capacity, and ΔT is the change in temperature. Since nitrogen gas is diatomic, its molar heat capacity at constant pressure is 5/2 R.
We need to rearrange the formula to find the change in temperature:
ΔT = Q / (n * C)
C = 5/2 R = 5/2 * 0.0821 L.atm/(K.mol) = 0.20525 L.atm/(K.mol)
ΔT = (26.6 kJ) / (54.69 mol * 0.20525 L.atm/(K.mol))
= 0.235 K
Finally, we can calculate the final temperature, T2:
T2 = T1 + ΔT = 297.15 K + 0.235 K
= 297.385 K
Converting back to Celsius:
T2 = 297.385 K - 273.15
= 24.235 °C
Therefore, the new temperature of the gas after adding 26.6 kJ of heat to the container will be approximately 24.235 °C.
To solve this problem, we can use the ideal gas law equation: 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 given temperature from °C to Kelvin. The Kelvin temperature scale is obtained by adding 273.15 to the Celsius temperature. So, the initial temperature of 24°C is 24 + 273.15 = 297.15 K.
Next, we can solve for the initial number of moles of nitrogen gas (n). Rearranging the ideal gas law equation to solve for n, we have n = PV / RT.
Given:
P = 3.5 atm
V = 440 L
R = 0.08206 L·atm/(mol·K)
T = 297.15 K
Substituting these values into the equation, we can calculate the initial number of moles:
n = (3.5 atm)(440 L) / (0.08206 L·atm/(mol·K))(297.15 K)
n ≈ 55.69 mol
Now, let's calculate the final temperature using the formula:
ΔT = Q / (n × Cv)
Where ΔT is the change in temperature, Q is the heat added, n is the number of moles, and Cv is the molar heat capacity at constant volume.
The molar heat capacity at constant volume (Cv) for nitrogen gas is approximately 20.79 J/(mol·K).
Given:
Q = 26.6 kJ = 26,600 J
n = 55.69 mol
Cv = 20.79 J/(mol·K)
Substituting these values into the formula, we can calculate the change in temperature:
ΔT = (26,600 J) / (55.69 mol × 20.79 J/(mol·K))
ΔT ≈ 23.43 K
Finally, we can determine the new temperature by adding the change in temperature to the initial temperature:
New temperature = 297.15 K + 23.43 K
New temperature ≈ 320.58 K
Converting the new temperature back to °C, we subtract 273.15:
New temperature ≈ 47.43°C
Therefore, the new temperature of the gas after adding 26.6 kJ of heat would be approximately 47.43°C.