`2000` mole of an ideal diatmic gas is enclosed in a vertica cylinder fitted with a piston and spring and spering as shown in the figeur. Initially, the spring is compressed by `5` cm and then the electric heater stats supplying energy to the gas at constant rate of `100 J//s` and due to conduction through walls of cylinder and radiation, `20 J//s` has been lost to surrounding.
`[k=1000N//m,g=10m//s^(2)` , Atmospheric pressure, `p_(0)=10^(5)N//m^(2)`, Cross-section area of piston `A=50cm^(2)` Mass of piston m `=1kg,R=8.3kJ//mol-K` ] Based on above information, answer the following question :
Increase in temperature of gas in `5` s is

Options

A. `6.9xx10^(-2)K`
B. `6.9xxK`
C. `83xx10^(-4)K`
D. `96xx10^(-4)K`

Answer

A

Explanation

At `t=0`, let `p` be the pressure of gas, then Free Body Diagram of piston would be as follows:

`impliesp_(0)A+Kx_(0)+mg=pA`
`impliesp=p_(0)+(kx_(0))/(A)+(mg)/(A)` brgt `impliesp=1.12 N//m^(2)`
From first law of thermodynamatics,
`dQ=dU+dW , dW=dQ-dU`
`=n(C_(p)-C_(v))dT=nRdT`
`W=nRxxDeltaT=114.3 J` Net heat supplied in time is,
`Q=int(100-20)dt=80t`
As heat has been supplied to the gas, it
expands under constant pressure, thus increasing the internal energy.
`dQ=nC_(p)dTimpliesQ=nC_(p)DeltaT`
`impliesDeltaT=(80xx5)/(2xx10^(3)xx(7)/(2)xx8.3)=6.9xx10^(-2)K`