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Multiple Choice
At what temperature in °C does 250.0 mL of argon gas at 25°C and 1.50 atm occupy a volume of 450.0 mL at a pressure of 1.90 atm, according to the Ideal Gas Law?
A
150°C
B
100°C
C
45°C
D
75°C
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Verified step by step guidance
1
Start by recalling the Ideal Gas Law, which is expressed as \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin.
Since the number of moles \( n \) and the gas constant \( R \) remain constant, you can use the combined gas law: \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \).
Convert the initial temperature from Celsius to Kelvin by adding 273.15 to the Celsius temperature: \( T_1 = 25 + 273.15 \).
Substitute the known values into the combined gas law equation: \( \frac{1.50 \text{ atm} \times 250.0 \text{ mL}}{T_1} = \frac{1.90 \text{ atm} \times 450.0 \text{ mL}}{T_2} \).
Solve for \( T_2 \) in Kelvin, and then convert it back to Celsius by subtracting 273.15 from the Kelvin temperature.