A sample of an gas has a volume of 2.25 L at 286 K and 1.13 atm. What is the pressure when the volume is 1.89 L and temperature is 302 K?

Why are the units of pressure in the RHS expression clearly as required?

To solve for the pressure when the volume is 1.89 L and the temperature is 302 K, we can use the combined gas law equation:

(P_1V_1/T_1 = P_2V_2/T_2)

Given: (V_1 = 2.25 , \text{L}) (T_1 = 286 , \text{K}) (P_1 = 1.13 , \text{atm}) (V_2 = 1.89 , \text{L}) (T_2 = 302 , \text{K})

Substituting the values into the equation:

(1.13 , \text{atm} \times 2.25 , \text{L} / 286 , \text{K} = P_2 \times 1.89 , \text{L} / 302 , \text{K})

Solving for (P_2):

(P_2 = (1.13 , \text{atm} \times 2.25 , \text{L} \times 302 , \text{K}) / (286 , \text{K} \times 1.89 , \text{L}))

(P_2 ≈ 1.55 , \text{atm})

To solve this problem, we can use the combined gas law, which states:

[ \frac{P_1 \cdot V_1}{T_1} = \frac{P_2 \cdot V_2}{T_2} ]

Where:

• ( P_1 ) and ( P_2 ) are the initial and final pressures, respectively.
• ( V_1 ) and ( V_2 ) are the initial and final volumes, respectively.
• ( T_1 ) and ( T_2 ) are the initial and final temperatures, respectively.

Given:

• ( V_1 = 2.25 , \text{L} )
• ( T_1 = 286 , \text{K} )
• ( P_1 = 1.13 , \text{atm} )
• ( V_2 = 1.89 , \text{L} )
• ( T_2 = 302 , \text{K} )

We can plug these values into the combined gas law equation to find ( P_2 ):

[ \frac{(1.13 , \text{atm}) \cdot (2.25 , \text{L})}{286 , \text{K}} = \frac{P_2 \cdot (1.89 , \text{L})}{302 , \text{K}} ]

Solving for ( P_2 ):

[ P_2 = \frac{(1.13 , \text{atm}) \cdot (2.25 , \text{L}) \cdot (302 , \text{K})}{(286 , \text{K}) \cdot (1.89 , \text{L})} ]

[ P_2 \approx \frac{(2.5425 , \text{atm} \cdot \text{L} \cdot \text{K})}{540.74 , \text{L} \cdot \text{K}} ]

[ P_2 \approx 0.0047 , \text{atm} ]

So, the pressure when the volume is ( 1.89 , \text{L} ) and the temperature is ( 302 , \text{K} ) is approximately ( 0.0047 , \text{atm} ).