Answer:
[tex]V_{2} = 75.6 mL[/tex]
Explanation:
To solve this problem we will use the Ideal Gas Law, which forms the following formula
[tex]PV = nRT[/tex]
Where
[tex]P[/tex] = Pressure of Gas
[tex]V[/tex] = Volume of Gas
[tex]n[/tex] = Moles of Gas
[tex]R[/tex] = Constant = 0.0821 atm L/mol K
[tex]T[/tex] = Temperature of Gas
There are two states to consider
State 1 is given
[tex]P_{1}[/tex] = 0.741 atm
[tex]V_{1}[/tex] = 112 mL
[tex]n_{1}[/tex] = unknown
[tex]R_[/tex] = 0.0821 atm L/mol K
[tex]T_{1}[/tex] = 300K
State 2 is based on Standard Room Temperature Pressure (STP) which are fixed values
[tex]P_{stp}[/tex] = 1 atm
[tex]V_{2}[/tex] = unknown
[tex]n_{2}[/tex] = unknown
[tex]R_[/tex] = 0.0821 atm L/mol K
[tex]T_{2}[/tex] = 273.15K
Since this is the same mass of gas it can be reasoned that the moles of gas are the same. We rearrange the Gas Law Formula to the following
[tex]n = \frac{PV}{RT}[/tex]
We input the values and equate the formulas for both states by substituting the value of n
[tex]\frac{P_{1}V_{1} }{RT_{1} } = \frac{P_{2}V_{2} }{RT_{2} } \\ \frac{(0.741)(112)}{(0.0821)(300)} = \frac{(1)(V_{2}}{(0.0821)(273.15)}[/tex]
Answer
[tex]V_{2} = 75.6mL[/tex]
