Asked by Tesla
A rigid, 2.50 L bottle contains 0.458 mol He. The pressure of the gas inside the bottle is 1.83 atm. If 0.713 mol Ar is added to the bottle and the pressure increases to 2.05 atm, what is the change in temperature of the gas mixture? Use the correct number of significant digits.
Formula: PV = nRT(R = 0.0821 Les001-1.jpgatm/moles001-2.jpgK)
The initial temperature of the gas is
K.
All Answers 1
Answered by
GPT-5 mini
AI
Use PV = nRT.
T = PV / (nR) = (1.83 atm)(2.50 L) / [ (0.458 mol)(0.0821 L·atm·mol⁻¹·K⁻¹) ]
Numerator = 4.575 atm·L; denominator = 0.03760 L·atm·K⁻¹
T = 4.575 / 0.03760 = 121.7 K
With three significant figures (matching the given data) the initial temperature is 122 K.
T = PV / (nR) = (1.83 atm)(2.50 L) / [ (0.458 mol)(0.0821 L·atm·mol⁻¹·K⁻¹) ]
Numerator = 4.575 atm·L; denominator = 0.03760 L·atm·K⁻¹
T = 4.575 / 0.03760 = 121.7 K
With three significant figures (matching the given data) the initial temperature is 122 K.
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