To find the volume of a sample of nitrogen gas at standard temperature and pressure (STP), given its initial conditions, we use the combined gas law:
[tex]\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \][/tex]
where:
- [tex]\( P_1 \)[/tex], [tex]\( V_1 \)[/tex], and [tex]\( T_1 \)[/tex] are the initial pressure, volume, and temperature respectively.
- [tex]\( P_2 \)[/tex], [tex]\( V_2 \)[/tex], and [tex]\( T_2 \)[/tex] are the final pressure, volume, and temperature respectively.
The initial conditions are:
- Volume, [tex]\( V_1 = 6.00 \)[/tex] liters
- Temperature, [tex]\( T_1 = 35^\circ \text{C} \)[/tex]
- Pressure, [tex]\( P_1 = 740 \)[/tex] torr
The standard conditions (STP) are:
- Temperature, [tex]\( T_2 = 0^\circ \text{C} = 273.15 \text{K} \)[/tex]
- Pressure, [tex]\( P_2 = 760 \)[/tex] torr
First, we need to convert the initial temperature from Celsius to Kelvin:
[tex]\[ T_1 (\text{K}) = T_1 (\text{C}) + 273.15 \][/tex]
[tex]\[ T_1 = 35 + 273.15 = 308.15 \text{K} \][/tex]
Next, we rearrange the combined gas law to solve for the final volume, [tex]\( V_2 \)[/tex]:
[tex]\[ V_2 = V_1 \left( \frac{P_1}{P_2} \right) \left( \frac{T_2}{T_1} \right) \][/tex]
Substitute the given values into the equation:
[tex]\[ V_2 = 6.00 \left( \frac{740}{760} \right) \left( \frac{273.15}{308.15} \right) \][/tex]
After simplifying the fractions and calculating, we find:
[tex]\[ V_2 \approx 5.178552823727337 \text{ liters} \][/tex]
Therefore, the volume of the nitrogen gas at STP is approximately 5.18 liters.
