A container holds 0.900g of oxygen at a pressure of 7.20 atmospheres .
(I) How much heat is required to increase the temperature to 116°?
(Il) HOw much will the temperature increase if this amount of heat energy is transferred to the gas at constant volume?
What kind of container? What is the specific heat of the container. What is ther specific heat of oxygen? What is the initial temperature of the container and oxygen?
To calculate the heat required to increase the temperature of the oxygen, we can use the formula:
Q = nCΔT
where:
Q is the heat required,
n is the number of moles of oxygen,
C is the molar heat capacity of oxygen at constant pressure, and
ΔT is the change in temperature.
To determine the number of moles of oxygen, 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.
Assuming the volume remains constant, we can rearrange the equation to solve for n:
n = PV / RT
Given:
P = 7.20 atmospheres
V (volume) = ?
R = 0.0821 L·atm/(mol·K)
T = 116°C + 273.15 (convert to Kelvin)
(I) First, let's find the number of moles of oxygen:
n = (7.20 atm) * V / (0.0821 L·atm/(mol·K)) * (116°C + 273.15 K) = 0.0316 mol
The molar heat capacity of oxygen at constant pressure (Cp) can be approximated as 5R/2:
Cp = 5 * (0.0821 L·atm/(mol·K)) / 2 = 0.2053 L·atm/(mol·K)
Now, we can calculate the heat required:
Q = (0.0316 mol) * (0.2053 L·atm/(mol·K)) * (116°C - 0°C) = 0.735 J
Therefore, the heat required to increase the temperature of the oxygen to 116°C is approximately 0.735 J.
(Il) To determine the increase in temperature when the same amount of heat energy is transferred to the gas at constant volume, we can use the specific heat capacity (Cv) for oxygen. Cv is approximately equal to Cp - R.
Cv = (5R/2) - R = (3R/2) = (3 * 0.0821 L·atm/(mol·K) / 2 = 0.12315 L·atm/(mol·K)
Now, we can use the formula:
Q = nCvΔT
We have already calculated Q as 0.735 J. Plugging in the values:
0.735 J = (0.0316 mol) * (0.12315 L·atm/(mol·K)) * ΔT
Solving for ΔT:
ΔT = 0.735 J / [(0.0316 mol) * (0.12315 L·atm/(mol·K))] ≈ 186 K
Therefore, the temperature will increase by approximately 186 Kelvin if the same amount of heat energy is transferred to the gas at constant volume.
To calculate the amount of heat required to increase the temperature of the oxygen in the container, we can use the formula:
q = m * C * ΔT
where:
q is the amount of heat (in J),
m is the mass of the oxygen (in kg),
C is the specific heat capacity of the oxygen (in J/kg·K),
ΔT is the change in temperature (in K).
(I) To find the amount of heat required to increase the temperature to 116°C, we need to convert the given values into the appropriate units.
1. Convert the mass from grams to kilograms:
m = 0.900g = 0.0009kg
2. Calculate the change in temperature in Kelvin:
ΔT = 116°C - 0°C = 116K
Now, we need to find the specific heat capacity of oxygen. The specific heat capacity depends on the conditions, and for oxygen at constant pressure, it is approximately 0.918 J/g·K.
3. Convert the specific heat capacity to J/kg·K:
C = 0.918 J/g·K * (1 kg / 1000 g) = 0.918 J/kg·K
Now, we can substitute the values into the formula:
q = (0.0009kg) * (0.918 J/kg·K) * (116K)
By evaluating this expression, we can find the amount of heat required to increase the temperature to 116°C.
(Il) To calculate the change in temperature if this amount of heat energy is transferred to the gas at constant volume, we need to use the equation:
q = m * C * ΔT
This equation, however, assumes constant pressure. Since the volume is held constant, we can assume the pressure remains the same.
We can rearrange the equation to solve for ΔT:
ΔT = q / (m * C)
Using the same values for mass (m) and specific heat capacity (C) from part (I), we can substitute them into the equation along with the value of heat energy (q).
ΔT = (q) / (0.0009kg * 0.918 J/kg·K)
Evaluating this expression will give us the change in temperature when the heat energy is transferred at constant volume.