Answer :
To determine the pressure of 5.0 moles of nitrogen gas in a 2.0-liter container at a temperature of 268 K, given the universal gas constant 0.0821 L-atm/(mol·K), we use the Ideal Gas Law:
[tex]\[ PV = nRT \][/tex]
Where:
- [tex]\( P \)[/tex] is the pressure in atmospheres (atm)
- [tex]\( V \)[/tex] is the volume in liters (L)
- [tex]\( n \)[/tex] is the number of moles (mol)
- [tex]\( R \)[/tex] is the universal gas constant, 0.0821 L-atm/(mol·K)
- [tex]\( T \)[/tex] is the temperature in Kelvin (K)
Rearrange the equation to solve for pressure [tex]\( P \)[/tex]:
[tex]\[ P = \frac{nRT}{V} \][/tex]
Substitute the given values into the equation:
[tex]\[ P = \frac{(5.0 \, \text{mol}) \times (0.0821 \, \text{L-atm/mol-K}) \times (268 \, \text{K})}{2.0 \, \text{L}} \][/tex]
Now, perform the multiplication and division:
[tex]\[ P = \frac{5.0 \times 0.0821 \times 268}{2.0} \][/tex]
[tex]\[ P = \frac{110.014}{2.0} \][/tex]
[tex]\[ P = 55.007 \, \text{atm} \][/tex]
Therefore, the pressure of the nitrogen gas in the container is [tex]\( 55.007 \, \text{atm} \)[/tex]. When rounded to two significant figures, the result is:
[tex]\[ P \approx 55 \, \text{atm} \][/tex]
Thus, the correct answer is:
D. 55 atm
[tex]\[ PV = nRT \][/tex]
Where:
- [tex]\( P \)[/tex] is the pressure in atmospheres (atm)
- [tex]\( V \)[/tex] is the volume in liters (L)
- [tex]\( n \)[/tex] is the number of moles (mol)
- [tex]\( R \)[/tex] is the universal gas constant, 0.0821 L-atm/(mol·K)
- [tex]\( T \)[/tex] is the temperature in Kelvin (K)
Rearrange the equation to solve for pressure [tex]\( P \)[/tex]:
[tex]\[ P = \frac{nRT}{V} \][/tex]
Substitute the given values into the equation:
[tex]\[ P = \frac{(5.0 \, \text{mol}) \times (0.0821 \, \text{L-atm/mol-K}) \times (268 \, \text{K})}{2.0 \, \text{L}} \][/tex]
Now, perform the multiplication and division:
[tex]\[ P = \frac{5.0 \times 0.0821 \times 268}{2.0} \][/tex]
[tex]\[ P = \frac{110.014}{2.0} \][/tex]
[tex]\[ P = 55.007 \, \text{atm} \][/tex]
Therefore, the pressure of the nitrogen gas in the container is [tex]\( 55.007 \, \text{atm} \)[/tex]. When rounded to two significant figures, the result is:
[tex]\[ P \approx 55 \, \text{atm} \][/tex]
Thus, the correct answer is:
D. 55 atm