Questions: If a gas at 30.12°C is at 0.86 atm and 2.3 L, what will be the temperature in Kelvin at 0.8 atm and 1.6 L? Provide answer to 2 decimal places. Do not include units.

Solution Steps

First, we need to convert the initial temperature from Celsius to Kelvin. The formula for this conversion is: \[ T(K) = T(^{\circ}C) + 273.15 \]

Given: \[ T_1 = 30.12^{\circ}C \]

So, \[ T_1(K) = 30.12 + 273.15 = 303.27 \, \text{K} \]

The combined gas law relates the pressure, volume, and temperature of a gas: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

Given: \[ P_1 = 0.86 \, \text{atm} \] \[ V_1 = 2.3 \, \text{L} \] \[ T_1 = 303.27 \, \text{K} \] \[ P_2 = 0.8 \, \text{atm} \] \[ V_2 = 1.6 \, \text{L} \]

We need to find \( T_2 \).

Rearrange the combined gas law to solve for \( T_2 \): \[ T_2 = \frac{P_2 V_2 T_1}{P_1 V_1} \]

Substitute the given values: \[ T_2 = \frac{(0.8 \, \text{atm}) (1.6 \, \text{L}) (303.27 \, \text{K})}{(0.86 \, \text{atm}) (2.3 \, \text{L})} \]

Perform the calculation: \[ T_2 = \frac{0.8 \times 1.6 \times 303.27}{0.86 \times 2.3} \] \[ T_2 = \frac{387.6864}{1.978} \] \[ T_2 \approx 196.00 \, \text{K} \]

Final Answer