the volume of a gas was originally 2.5 L; it's pressurewas 104kpa and its temperatire was 270K. the volume of the gas expanded to 5.3 L and its pressure decreased to 95 kpa. What is the temperature of the gas?
(P1V1/T1) (P2V2/T2)
T must be in kelvin.
To find the temperature of the gas, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant (8.314 J/(mol·K))
T = temperature in Kelvin
First, we need to find the number of moles of the gas using the initial conditions.
Initial conditions:
V1 = 2.5 L
P1 = 104 kPa
T1 = 270 K
Using the ideal gas law equation, we can rearrange it to solve for n:
n = (P1 * V1) / (R * T1)
Substituting the given values:
n = (104 kPa * 2.5 L) / (8.314 J/(mol·K) * 270 K)
n ≈ 0.0112 mol
Next, we can use the final conditions to find the temperature of the gas.
Final conditions:
V2 = 5.3 L
P2 = 95 kPa
Using the ideal gas law equation:
T2 = (P2 * V2) / (n * R)
Substituting the known values:
T2 = (95 kPa * 5.3 L) / (0.0112 mol * 8.314 J/(mol·K))
T2 ≈ 226.6 K
Therefore, the temperature of the gas is approximately 226.6 K.
To find the temperature of the gas, we can use the combined gas law. The combined gas law states:
(P1 * V1) / T1 = (P2 * V2) / T2
Where:
P1 = Initial pressure
V1 = Initial volume
T1 = Initial temperature
P2 = Final pressure
V2 = Final volume
T2 = Final temperature
In this case, we are given the following values:
P1 = 104 kPa
V1 = 2.5 L
T1 = 270 K
P2 = 95 kPa
V2 = 5.3 L
Now we can rearrange the equation to solve for T2:
T2 = (P2 * V2 * T1) / (P1 * V1)
Substituting the given values:
T2 = (95 kPa * 5.3 L * 270 K) / (104 kPa * 2.5 L)
T2 ≈ 405 K
Therefore, the temperature of the gas is approximately 405 K.
For an ideal gas,
(P1*V1)/(T1) = (P2*V2)/(T2)
where
P1, V1 and T1 are the pressure, volume and temp at initial conditions, and
P2, V2 and T2 are the final conditions
Substituting,
(104*2.5)/270 = (95*5.3)/(T2)
now solve for T2.