2.00 mol of a monatomic ideal gas `(U=1.5nRT)` is enclosed in an adiabatic, fixed , vertical cylinder fitted with a smooth, light adiabatic piston.the piston is connected to a vertical spring of spring constant `200 N m^(-1)` as shown in .the area of cross section of the cylinder is `20.0 cm^(2)`. initially, the spring is at its natural length and the temperature of the gas is at its natural length and the temperature of the gas is 300 K. the atmosphere pressure is 100 kPa. the gas is heated sloewly for some time by means of an electric heater so as to move the poston up through 10 cm. find (a) the work done by the gas (b) the final temperature of the gas and (c ) the heat supplied by the heater.

Explanation

(a) the force by the gas on the piston is
`F=p_(0)A+ks`
Where `p_(0)=100 kPa` is the atmospheric pressure, `A=20 cm^(2)` is the area of cross section, `k=200 Nm^(1)` is the spring constant and x is the compression of the spring. The work done by the gas as the piston moves through`l=10 cm is
`W=int_(0)^(l) Fdx`
`=p_(0)Al+(1)/(2)kl^(2)`
=(100xx10^(3)Pa)xx(20xx10^(-4)m^(2))xx(10xx10^(-2)m)`
`+(1)/(2)(200Nm^(-1))xx(100xx10^(-4)m^(2))`
`=20 J+1J=21 J`.
(b) The initial tempreture is `T_(1)=300 K`. Let the final tempreture be `T_(2)` We have
`nRT_(1)=p_(0)V_(0)`
and `nRT_(2)=p_V_(2)=(p_(0)+(kl)/A)(V_(0)+Al)`
`=nRT_(1)+p_(0)Al+kl^(2)+(klnRT_(1))`/(Ap_(0))`
or, `T_(2)=T_(1)+(p_(0)Al+kl^(2))/(nR)+(klT_(1))/(Ap_(0))`
`=(300 K)+(20 J+2 J)/((2.0 mol)(8.3 JK^(-1) mol^(-1))`
`+((200 Nm^(-1))xx(10xx10^(-2)m)xx(300 K))/((20xx10^(4)m^(2))xx(100xx10^(3)Pa))`
`=300 K+1.325 K+30 K`
`=331 K`.
(c ) the internal energy is `U=1.5nRT`.
the change in internal energy is
`DeltaU=1.5nRDeltaT`
`=1.5xx(2.00 mol)xx(8.3 J K^(-1) mol^(-1)xx(31 K)`
`=772 J
From the first law,
`DeltaQ=DeltaU+DeltaW`
`772 J+21 J=793 J`.