MEDIUMMCQ SINGLEJEE Advanced ChemistryPhysical ChemistryGaseous statePrevious Year Questions
The temperature at which 28 g of N2 will occupy a volume of l0.0 L at 2.46 atm is (RPMT 2010)
A.299.6 K (b) 0°C✓ Correct
C.273 K (d) 10oC
Explanation
28g of N2 is equal to 1 mole of N2 Applying, ideal gas equation pV = nRT, 2.46 × 10 = 1 × 0.0821 × T T = 299.6 K
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