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Exercises · 5.11

Q.A student forgot to add the reaction mixture to the round bottomed flask at 27 °C but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477 °C. What fraction of air would have been expelled out?

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Step 1 – Identify the constraint

The flask is open to the air, so its volume VV and (atmospheric) pressure pp both stay CONSTANT as it's heated; only the amount of gas nn inside decreases (air is pushed out) as TT rises. From pV=nRTpV=nRT, with pVpV fixed:

n1T1=n2T2=pVR=constantn_1T_1 = n_2T_2 = \frac{pV}{R} = \text{constant}

Step 2 – Convert temperatures to kelvin

T1=27+273=300 K,T2=477+273=750 KT_1 = 27+273 = 300\ \text{K}, \qquad T_2 = 477+273 = 750\ \text{K}

Step 3 – Find the ratio of moles remaining …

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