Question

what is the number of moles of gas contained in a 3.0l vessel at 300k with a pressure of 1.50 atm? (r =.0821 latm/molk)
a).18 mol
b).45 mol
c) 1.18 mol
d) 4.50 mol

1. if the pressure exerted by a gas at 25c in a volume of.044l is 3.81 atm, how many moles of gas are present? (r =.0821 latm/molk)

a).0069 mol
b).69 mol
c) 6.9 mol
d) 69 mol

1. there are 6.2 liters of an ideal gas is contained at 3.0 atm and 37 k. how many moles of this gas are present?

Explanation:

Step1: Recall ideal - gas law

The ideal - gas law is $PV = nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the ideal - gas constant ($R=0.0821\ L\cdot atm/mol\cdot K$), and $T$ is temperature in Kelvin. We need to solve for $n$, so $n=\frac{PV}{RT}$.

Step2: Solve the first problem

Given $V = 3.0\ L$, $T = 300\ K$, $P = 1.50\ atm$, and $R=0.0821\ L\cdot atm/mol\cdot K$.
Substitute the values into the formula: $n=\frac{1.50\ atm\times3.0\ L}{0.0821\ L\cdot atm/mol\cdot K\times300\ K}=\frac{4.5}{24.63}\approx0.18\ mol$.

Step3: Solve the second problem

First, convert the temperature from Celsius to Kelvin. $T=(25 + 273)K=298\ K$, $V = 0.044\ L$, $P = 3.81\ atm$, $R = 0.0821\ L\cdot atm/mol\cdot K$.
Substitute into $n=\frac{PV}{RT}$: $n=\frac{3.81\ atm\times0.044\ L}{0.0821\ L\cdot atm/mol\cdot K\times298\ K}=\frac{0.16764}{24.4658}\approx0.0069\ mol$.

Step4: Solve the third problem

Given $V = 6.2\ L$, $P = 3.0\ atm$, $T = 37\ K$, $R = 0.0821\ L\cdot atm/mol\cdot K$.
Substitute into $n=\frac{PV}{RT}$: $n=\frac{3.0\ atm\times6.2\ L}{0.0821\ L\cdot atm/mol\cdot K\times37\ K}=\frac{18.6}{3.0377}\approx6.12\ mol$.

Answer:

1. First problem: None of the given options are correct.
2. Second problem: A. $0.0069\ mol$
3. Third problem: None of the given options are correct.