Air is filled in an open container at 60°C temperature. The temperature at which 1/4 the part of the air goes outside is?

Question

Norman Manchu · Answer

Using the ideal equation we can eliminate the constants and take help of the quantities that are varying Pressures remains constant as in both cases the vessel is open to the atmosphere thus P=1atm The vessel in both the cases is same Hence the volume remains same as well R is universal gas constant whose value of course remains constant Thus PV|R is constant for both the cases and equal to n1T1 and n2T2 for respective situations Equate them n1 would be n and T1 would be 333K (Never put temperature in Celsius here) n2=n-n|4=3n|4(n|4 is the amount or quantity of gas lost) T2 has to be calculated and it would be the answer=171degree'Celsius

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Air is filled in an open container at 60°C temperature. The temperature at which 1/4 the part of the air goes outside is?

Air is filled in an open container at 60°C temperature. The temperature at which 1/4 the part of the air goes outside is?

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