Calculate the volume at s.t.p. occupied by a gas 'Q' originally occupying 153.7cc at 287K and 750mm pressure [vapor pressure
of at 287K is 12mm of Hg]
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To calculate the volume occupied by gas 'Q' at standard temperature and pressure (STP), you will need to use the ideal gas law equation, which states:
PV = nRT,
where:
P is the pressure of the gas,
V is the volume of the gas,
n is the number of moles of gas,
R is the ideal gas constant (0.0821 L.atm/mol.K), and
T is the temperature of the gas in kelvin.
Given information:
Initial volume (V1) = 153.7 cc = 0.1537 L,
Initial temperature (T1) = 287 K,
Initial pressure (P1) = 750 mmHg,
Vapor pressure at 287 K (P_vapor) = 12 mmHg.
First, we need to convert the initial pressure and vapor pressure to atmospheres (atm):
P1 = 750 mmHg * (1 atm / 760 mmHg) = 0.9868 atm,
P_vapor = 12 mmHg * (1 atm / 760 mmHg) = 0.0158 atm.
Next, we can rearrange the ideal gas law equation to solve for the final volume (V2):
V2 = (n * R * T2) / P2,
where T2 is the final temperature, and P2 is the final pressure. Since we are dealing with the same gas (Q) at the same conditions of temperature and pressure, we can assume that n (number of moles) remains constant.
Now, at STP, the temperature is 273 K and the pressure is 1 atm.
Substituting the values into the equation, we can solve for V2:
V2 = (n * R * 273 K) / (1 atm).
Since we assume the number of moles of gas (n) remains constant, we can write:
V1 / P1 = V2 / P2.
Rearranging the equation to solve for V2:
V2 = (P2 * V1 * P1) / P_vapor.
Substituting the given values into the equation:
V2 = (1 atm * 0.1537 L * 0.9868 atm) / 0.0158 atm.
Calculating the value:
V2 = 9.594 L.
Therefore, the volume occupied by gas 'Q' at STP is approximately 9.594 L.