Find the temperature at which volume and pressure of 28gm nitrogen gas become 10dm3 and 2.46 atmosphere respectively
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A)
W = 28g
P = 2.46 atm
V = 10 litres
m = 28
∴PV=wmRT (R = 0.0821 litre atm/K/mol)
T=mPVw×R
T=28×2.46×1028×0.0821
=688.82.2988
= 299.64 K
Hence answer is (b)
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