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Consider a vertical tube open at bothh ends. The cube consists of two parts, each of different cross-sections and each parth having a piston which can move smoothly in respective tubes. The two pistons are joined together by an inextensible wire. The combined mass of the two piston is 5 kg and area of cross-section of the upper piston is 10 cm^(2) greater than that of the lower piston. Amount of gas enclosed by the pistons is one mole. When the gas is heated slowly, pistons move by 50 cm. Find rise in the temperature of the gas, in the form 25X//R K where R is universal gas constant. Use g=10m//s^(2) and outside pressure =10^(5)N//m^(2)). Fill value of X in the answer sheet.

Answer»

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Solution :DUE to the HEATING pressure inside in not changed. Let inside pressure be `rho`. Then for equilibrium of the system.

`P(A_(1)-A_(2))=P_(0)(A_(1)-A_(2))+(m_(1)+m_(2))g`
`impliesP/_\V=(P_(0)/_\A+mg)L`
`L` is displacement of the piston.
`P./_\V=nR/_\T`
`/_\T=(P/_\V)/(nR)=((P_(0)/_\V+mg)L)/(nR)`
`=((10^(5)P_(a)xx10^(-3)m^(2)+5xx10)(50xx10^(-2)))/(1xxR)=75/R K`


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