> #gas-laws , #thermodynamics , #physics > A 6.7l volume of air,initially at 23.c and 0.98atm, is compressed to 2.7l while heated to 125 .c. which is the final pressure?
Jan 12

# A 6.7l volume of air,initially at 23.c and 0.98atm, is compressed to 2.7l while heated to 125 .c. which is the final pressure?

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Jan 12
To solve this problem, we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin

First, we need to convert the initial and final temperatures from Celsius to Kelvin:

Initial temperature (T1) = 23°C + 273.15 = 296.15 K
Final temperature (T2) = 125°C + 273.15 = 398.15 K

Next, we can calculate the initial number of moles (n1) using the ideal gas law equation:

P1V1 = n1RT1

n1 = (P1V1) / (RT1)

Given:
P1 = 0.98 atm
V1 = 6.7 L
T1 = 296.15 K
R = 0.0821 L·atm/(mol·K)

n1 = (0.98 atm * 6.7 L) / (0.0821 L·atm/(mol·K) * 296.15 K)
n1 ≈ 0.285 mol

Now, we can use the ideal gas law equation to find the final pressure (P2):

P2V2 = n2RT2

Since the number of moles (n2) remains constant, we can rewrite the equation as:

P2 = (n1 * R * T2) / V2

Given:
n1 = 0.285 mol
R = 0.0821 L·atm/(mol·K)
T2 = 398.15 K
V2 = 2.7 L

P2 = (0.285 mol * 0.0821 L·atm/(mol·K) * 398.15 K) / 2.7 L
P2 ≈ 10.5 atm

Therefore, the final pressure is approximately 10.5 atm.
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