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Solution :

Given that initial pressure, volume and temperature of gas is <br> `P_1=1.6xx10^6 N//m^2` <br> `V_1=0.0083m^3` <br> `T_1`=300 K <br> From gas law we can find the number of moles of gas in the container as <br> `n=(PV)/(RT)=(1.6xx10^6xx0.0083)/(8.314xx300)` <br> `=16/3`mole <br> It is given that molar specific heat of gas at constant pressure is <br> `C_P=(5R)/2` <br> Thus gas is monoatomic, hence its molar specific heat at constant volume is given as <br> `C_V=(3R)/2` <br> As gas is heated in a closed vessel i.e. at constant volume, if its temperature is raised from `T_1` to `T_2` then, we have heat supplied to the gas is <br> `Q=n C_V(T_2-T_1)` <br> or `2.49xx10^4=16/3xx3/2R(T_2-300)` <br> or `T_2=300+(3xx2xx2.49xx10^4)/(16xx3xx8.314)` <br> `=300+374.36` <br> =674.36 K <br> For constant volume process , we have <br> `P_2/P_1=T_2/T_1` <br> Thus `P_2=T_2/T_1xxP_1` <br> `=674.36/300xx1.6xx10^6` <br> `=3.6xx10^6 N//m^2`