Asked by vianna trigueros on May 15, 2024
Verified
If a gas at 25.0 atm and 10.0L at a temperature of 15.0°C is pressurized to 30 atm and heated to 20°C, what is its new volume?
A) 11.1 L
B) 333 L
C) 16.7 L
D) There is not enough information to tell.
Atm
A unit of pressure defined as atmospheric pressure, equivalent to the pressure exerted by the Earth's atmosphere at sea level.
Volume
The quantity of three-dimensional space occupied by a substance or item.
°C
A unit of temperature on the Celsius scale, a temperature scale originally known as the centigrade scale, with 0°C denoting the freezing point of water and 100°C its boiling point at sea level.
- Expound on how the ideal gas law governs the relationship of temperature, pressure, volume, and moles in gases.
Verified Answer
EF
Erica FrancoMay 20, 2024
Final Answer :
A
Explanation :
Using the combined gas law, P1V1/T1=P2V2/T2P_1V_1/T_1 = P_2V_2/T_2P1V1/T1=P2V2/T2 , where PPP is pressure, VVV is volume, and TTT is temperature in Kelvin. Convert temperatures to Kelvin: T1=15.0°C+273.15=288.15KT_1 = 15.0°C + 273.15 = 288.15KT1=15.0°C+273.15=288.15K and T2=20°C+273.15=293.15KT_2 = 20°C + 273.15 = 293.15KT2=20°C+273.15=293.15K . Solve for V2V_2V2 : V2=P1V1T2P2T1=25.0×10.0×293.1530×288.15≈11.1LV_2 = \frac{P_1V_1T_2}{P_2T_1} = \frac{25.0 \times 10.0 \times 293.15}{30 \times 288.15} \approx 11.1LV2=P2T1P1V1T2=30×288.1525.0×10.0×293.15≈11.1L .
Learning Objectives
- Expound on how the ideal gas law governs the relationship of temperature, pressure, volume, and moles in gases.