Questions: Solve the application problem. The volume V of an ideal gas varies directly with the temperature T and inversely with the pressure P. A cylinder contains oxygen at a temperature of 320 degrees K and a pressure of 17 atmospheres in a volume of 140 liters. Find the pressure if the volume is decreased to 130 liters and the temperature is increased to 340 degrees K. (Round your answer to one decimal place.) 19.3 x atmospheres

Solution Steps

The problem states that the volume \( V \) of an ideal gas varies directly with the temperature \( T \) and inversely with the pressure \( P \). This relationship can be expressed as:

\[ V = k \frac{T}{P} \]

where \( k \) is a constant.

Using the initial conditions provided, we can find the constant \( k \). We have:

• Initial volume \( V_1 = 140 \) liters
• Initial temperature \( T_1 = 320 \) K
• Initial pressure \( P_1 = 17 \) atmospheres

Substitute these values into the equation:

\[ 140 = k \frac{320}{17} \]

Solving for \( k \):

\[ k = 140 \times \frac{17}{320} = \frac{2380}{320} = 7.4375 \]

Now, use the constant \( k \) to find the new pressure \( P_2 \) when the volume is decreased to 130 liters and the temperature is increased to 340 K.

• New volume \( V_2 = 130 \) liters
• New temperature \( T_2 = 340 \) K

Substitute these values into the equation:

\[ 130 = 7.4375 \frac{340}{P_2} \]

Solving for \( P_2 \):

\[ P_2 = 7.4375 \times \frac{340}{130} \]

\[ P_2 = \frac{2528.75}{130} = 19.4529 \]

Rounding to one decimal place, \( P_2 = 19.5 \) atmospheres.

Final Answer