Question #96245

if a 15.0 celcius tempature increase causes a 10.0 percent increase in the volume of a 832 ml sample of He(g) while the gas pressure is held constant what was the oridinal temperature?

Expert's answer

Use Charle's law



V1T1=V2T2\frac{V_1}{T_1}=\frac{V_2}{T_2}

T2=15+T1T_2 = 15+T_1

V1=832cm3V_1 = 832 cm^3

V2=832+0.1×832=915.2cm3V_2 = 832 + 0.1\times832=915.2 cm^3



832T1=915.2T1+15\frac{832}{T_1}=\frac{915.2}{T_1+15}

T1(K)=150KT_1 (K) = 150 K

T2(∘C)=150−273=−123=−123∘CT_2 (^\circ C) = 150 -273 = -123 =-123^\circ C


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