Final answer:
The argon gas sample's temperature must be increased to approximately 72.28°C in order to decrease its volume to 5.78 L and increase its pressure to 1.50 atm. This is calculated using the combined gas law formula which describes the behavior of gases in a closed container.
Step-by-step explanation:
This question involves using the combined gas law equation P₁V₁/T₁ = P₂V₂/T₂, where P is pressure, V is volume, and T is temperature (on the Kelvin scale). Initially, we have P₁ = 1.19 atm, V₁ = 6.86 L, and T₁ = 53.0 + 273.15 (converting Celsius to Kelvin).
We want to find T₂ when P₂ = 1.50 atm and V₂ = 5.78 L. After plugging in the known values and cross-multiplying, we get T₂ = P₂V₂T₁ / P₁V₁, and by calculation, T₂ comes out to be around 345.4K approximately. However, since the question asks for the result in Celsius, we subtract 273.15 from our answer, giving T₂ = 72.28℃.
Main answer:
To decrease the volume of the argon gas sample to 5.78 L while increasing its pressure to 1.50 atm, the temperature must be increased to approximately 72.28°C.
Step-by-step explanation:
The gas laws describe how changes in temperature, pressure, and volume behave in a closed container of gas. The combined gas law allows us to compare the initial and final states of the gas sample.
Conclusion:
When changing the conditions of a container of gas, it is important to understand the relationships between pressure, volume, and temperature to predict and calculate changes in state.
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