Answer :
At Standard Temperature and Pressure (STP) the volume of the gas at is 216.1 ml.
A sample of ideal gas occupies 208ml at 36.2 degree Celsius and 704 torr what is the volume at stp
For a sample of ideal gas, the relationship between volume, pressure, and temperature is given by the Ideal Gas Law:
PV = nRT
Where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the universal gas constant, and T is the temperature of the gas.
At STP (Standard Temperature and Pressure), the pressure of the gas is 1 atm and the temperature of the gas is 0°C (273 K). Therefore:
P1 = 704 torr
V1 = 208 mL
T1 = 36.2°C = 309.35 K
P2 = 1 atm
V2 = ?
T2 = 0°C = 273 K
To find V2, we can use the following equation:
V2 = V1(P2/P1)(T1/T2)
Plugging in the given values:
V2 = 208 mL (1 atm/704 torr) (309.35 K/273 K)
V2 = 208 mL (0.939) (1.132)
V2 = 216.1 mL
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Using the combined gas law, we determined that the volume of the gas at STP is approximately 171 ml. We converted the given initial conditions to standard temperature and pressure to find the new volume. By substituting values into the combined gas law formula, we performed the calculation to arrive at the final volume.
To find the volume of a gas at STP (Standard Temperature and Pressure), we can use the combined gas law. The combined gas law formula is:
[tex]\frac{P_1V_1}{T_1}= \frac{P_2V_2}{T_2}[/tex]
Given:
- [tex]P_1[/tex] = 704 torr (0.926 atm, as 1 atm = 760 torr)
- [tex]V_1[/tex] = 208 ml (0.208 L)
- [tex]T_1[/tex] = 36.2°C (309.35 K, as T(K) = T(°C) + 273.15)
- Standard Temperature ([tex]T_2[/tex]) = 0°C or 273.15 K
- Standard Pressure ([tex]P_2[/tex]) = 1 atm
Rearranging the combined gas law equation to solve for V2 gives:
[tex]V_2[/tex] = [tex]\frac{P_1V_1T_2}{T_1}[/tex]
Substituting the given values:
[tex]V_2[/tex] = (0.926 atm * 0.208 L * 273.15 K) / (309.35 K * 1 atm)
Calculating this yields:
[tex]V_2[/tex] ≈ 0.171 L or 171 ml
Therefore, the volume of the gas at STP is approximately 171 ml.
