At s.t.p, a mole of chlorine occupies 22.4dm3. If the temperature is changed to 35°C and the pressure to 750mHg the of the new volume will be [Cl = 35.5]
\(\frac{760\times 22.4\times 308}{750\times 273\times 35.5} dm^3 \)
\(\frac{760\times 22.4\times 308}{750\times 273}dm^3 \)
\(\frac{750\times 22.4\times 35.5}{760\times 273}dm^3\)
\(\frac{760\times 22.4\times 35.5}{750\times 22.3\times 273}dm^3 \)
\(\frac{760\times 22.4\times 35.5}{750\times 308}dm^3 \)
Explanation
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Discussions (7)
its quite simple...
At S.T.P
p1=760mmhg
T1=273k
V1=22.4dm³
P2=750mmHg
T2= (35°+273°)=308k
V2 = ?
(p1v1)/T1 = (p2v2)/T2
v2 = (p1v1T2)/p2T1
now kindly substitute the letters for they figures
V2 = (760×22.4×308)/750×273
jamb was even kind enough to cut our solving short and save time
Owk now I understand,they used step for the first pressure and temp. then they did p1v1/t2=p2v2/t2 den cross multiply..... of u look closely you would see the x
