SP 1: 2.085g H2 produces a pressure of 1.015 atm in a 5.00L container at -211.76°C. What will the temperature (in °C) be if an additional 2.099g H2 are added to the container and t... SP 1: 2.085g H2 produces a pressure of 1.015 atm in a 5.00L container at -211.76°C. What will the temperature (in °C) be if an additional 2.099g H2 are added to the container and the pressure increases to 3.015atm? SP 2: A 10.8g sample of gas occupies 22.914L at STP. What is the weight of this gas? Read more

Steps to Solve

1. Identifying Given Values

   For SP 1:

   • Initial mass of H₂: $m_1 = 2.085 , \text{g}$
   • Additional mass of H₂: $m_2 = 2.099 , \text{g}$
   • Final mass: $m = m_1 + m_2 = 4.134 , \text{g}$
   • Initial pressure: $P_1 = 1.015 , \text{atm}$
   • Final pressure: $P_2 = 3.015 , \text{atm}$
   • Volume: $V = 5.000 , \text{L}$
   • Initial temperature: $T_1 = -211.76 , \text{°C} = -211.76 + 273.15 = 61.39 , \text{K}$

2. Using the Ideal Gas Law

   We use the ideal gas law for two states given by: $$ P_1 V = n_1 R T_1 $$ $$ P_2 V = n_2 R T_2 $$

   This helps us relate the two temperatures.

3. Converting to Moles

   The number of moles, $n$, can be calculated using: $$ n = \frac{m}{M} $$ where $M$ is the molar mass of hydrogen (about $2.02 , \text{g/mol}$).

   So, $$ n_1 = \frac{2.085 , \text{g}}{2.02 , \text{g/mol}}, \quad n_2 = \frac{4.134 , \text{g}}{2.02 , \text{g/mol}} $$

4. Rearranging for Final Temperature

   Rearranging the ideal gas law for the second state: $$ T_2 = \frac{P_2 m_1 T_1}{P_1 m_2} $$

   Substitute the previously calculated values: $$ T_2 = \frac{3.015 , \text{atm} \cdot (2.085 , \text{g}) \cdot (61.39 , \text{K})}{1.015 , \text{atm} \cdot (4.134 , \text{g})} $$

5. Final Calculation

   Calculate $T_2$ with the above values:

   • First calculate the numerator: $3.015 \cdot 2.085 \cdot 61.39$
   • Then calculate the denominator: $1.015 \cdot 4.134$
   • Finally divide the two results to get $T_2$.