Respuesta :
Answer: The volume of argon gas at STP is 57.4 mL
Explanation:
STP conditions are:
Pressure of the gas = 1 atm = 760 torr
Temperature of the gas = 273 K
To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law. The equation follows:
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
where,
[tex]P_1,V_1\text{ and }T_1[/tex] are the initial pressure, volume and temperature of the gas
[tex]P_2,V_2\text{ and }T_2[/tex] are the final pressure, volume and temperature of the gas
We are given:
[tex]P_1=725torr\\V_1=65.0mL\\T_1=22^oC=(22+273)K=295K\\P_2=760torr\\V_2=?\\T_2=273K[/tex]
Putting values in above equation, we get:
[tex]\frac{725torr\times 65.0mL}{295K}=\frac{760torr\times V_2}{273K}\\\\V_2=57.4mL[/tex]
Hence, the volume of argon gas at STP is 57.4 mL
Explanation:
The given data is as follows.
[tex]V_{1}[/tex] = 65.0 mL = 0.065 L (as 1 ml = 0.001 L),
[tex]T_{1}[/tex] = [tex]22.0^{o}C[/tex] = (22 + 273) K = 295 K,
[tex]P_{1}[/tex] = 725 torr = 0.954 atm (as 1 torr = 0.00131579 atm),
[tex]V_{2}[/tex] = ?, [tex]T_{2}[/tex] = 273 K,
[tex]P_{2}[/tex] = 1 atm
And, according to ideal gas equation,
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]
[tex]\frac{0.954 atm \times 0.065 L}{295 K} = \frac{1 atm \times V_{2}}{273 K}[/tex]
[tex]V_{2}[/tex] = 0.0574 L
As, 1 L = 1000 ml. So, 0.0574 L = 57.4 ml.
Thus, we can conclude that the volume of this Ar gas at STP is 57.4 L.