| Given conditions |
Final conditions |
V1 = 35 ml
P1 = 758 mm
T1 = 6+273
= 279 K |
V2 = ?
P2 = 760 mm
T2 = 0 + 273
= 273 K |
By applying ideal gas equation, we have
\(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\)
\(\frac{758\times35}{279}\) = \(\frac{760\times V_2}{273}\)
\(\therefore\) V2 = \(\frac{758\times35}{279}\) x \(\frac{273}{760}\)
= 34∙16 ml
Volume of chlorine gas at NIP = 34∙16 ml.
