the volume at s.t.p. occupied by a gas q originally occupying 153.7 cc at 287 k and 750mm pressure, vapour pressure of gas q at 287k is 12 mm of hg
Explanation:
Given data:
Volume of gas = 153.7 cm^3
Temperature "T" = 287 K
Pressure "P" = 750 mm
P1V1/ T1 = P2V2 / T2
We have p1 = 750 + vapour pressure of gas at 287 K
= (750 + 12) mm of Hg = 762 mm of Hg
V1 = 153.7 cm^3 and T1 = 287 K
p2 = 760 mm of Hg , T2 = 273 K
(As STP conditions are 1 atm pressure which is equal to 760 mm of Hg and 273K temperature)
And V2 = ?
Therefore V2 = (p1V1) / T1 X (T2 / p2)
= (762 X 153.7 X 273) / ( 287 X 760)
= 146.6 cm^3
Hence the second Volume is 146.6 cm^3.
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To find the volume occupied by gas q at STP (Standard Temperature and Pressure), we need to use the ideal gas law equation:
PV = nRT
where:
P = pressure
V = volume
n = number of moles
R = the ideal gas constant
T = temperature
First, we need to calculate the number of moles of gas q:
We can use the equation PV = nRT to calculate the number of moles:
n = PV / RT
Given:
P = 750 mmHg
V = 153.7 cc
T = 287 K
R = 0.0821 L·atm/mol·K (ideal gas constant)
Converting the given values:
P = 750 mmHg × (1 atm / 760 mmHg) = 0.987 atm
V = 153.7 cc × (1 L / 1000 cc) = 0.1537 L
Now, substitute the values into the equation to find the number of moles (n):
n = (0.987 atm) × (0.1537 L) / (0.0821 L·atm/mol·K × 287 K)
n ≈ 0.007694 moles
Now, we can determine the volume occupied by gas q at STP. At STP, the temperature is 273.15 K, and the pressure is 1 atm.
Using the ideal gas law equation again (PV = nRT), we can solve for V:
V = (nRT) / P
Let's substitute the values into the equation:
V = (0.007694 moles) × (0.0821 L·atm/mol·K × 273.15 K) / 1 atm
V ≈ 0.0164 L
Converting the volume to cc:
V(STP) = 0.0164 L × (1000 cc / 1 L)
V(STP) ≈ 16.4 cc
Therefore, the volume occupied by gas q at STP is approximately 16.4 cc.
To find the volume of a gas at STP (Standard Temperature and Pressure), we need to use the ideal gas law equation. The ideal gas law is stated as:
PV = nRT
Where:
P = Pressure
V = Volume
n = Number of moles of gas
R = Ideal gas constant
T = Temperature
The ideal gas constant (R) is 0.0821 L*atm/(mol*K), and the STP conditions are defined as 1 atmosphere (atm) of pressure and 273.15 Kelvin (K) of temperature.
In this case, you have the initial volume (153.7 cc) at a certain temperature (287 K) and pressure (750 mm). However, to use the ideal gas law, we need to convert the given values into SI units.
1 cc = 1 mL = 0.001 L
1 mmHg = 0.00131579 atm
Converting the given values:
Initial Volume (V) = 153.7 cc * 0.001 L/cc = 0.1537 L
Pressure (P) = 750 mmHg * 0.00131579 atm/mmHg ≈ 0.9868 atm
Temperature (T) = 287 K
Now we can substitute these values into the ideal gas law equation:
(0.9868 atm) * (0.1537 L) = n * (0.0821 L*atm/(mol*K)) * (287 K)
Simplifying the equation:
0.151 (mol*atm) = 0.0821 (mol*K) * n
To isolate n, divide both sides of the equation by (0.0821 (mol*K)):
n = 0.151 (mol*atm) / 0.0821 (mol*K)
n ≈ 1.84
Now we know the number of moles (n) of the gas. To find the volume at STP, we can rearrange the ideal gas law equation:
P * V = n * R * T
To find the volume (V) at STP, we substitute the known values:
P = 1 atm
T = 273.15 K
n = 1.84 moles
R = 0.0821 L*atm/(mol*K)
(1 atm) * V = (1.84 moles) * (0.0821 L*atm/(mol*K)) * (273.15 K)
Simplifying the equation:
V ≈ (1.84 * 0.0821 * 273.15) / 1
V ≈ 38.14 L
Therefore, the volume of the gas q at STP conditions (1 atm and 273.15 K) is approximately 38.14 L.