The temperature of a gas contained in a closed vessel increases by 1 degree Celsius when pressure of the gas is increased by 1%. The initial temperature of the gas is ?

Solution: according to Boyle's law, PV=nRT
Let, initially the gas had pressure=P, volume=V, No. of moles=n and temperature=T degree Celsius
Therefore, the initial temperature in Kelvin scale is T + 273degree
After increasing 1 degree Celsius the pressure increased by 1%. Hence, the new temp. is=T+1+273
                                                                                                                                             =T+274K
and new pressure=(101/100)*P
In summery, PV=nR(T+273)So, PV/nR=T+273--------------1
                     (101/100)*PV=nR(T+274),So (101/100)*PV/nR=T+274-------------2
substituting PV/nR value from equation 1 in equation 2,
(101/100)*(T+273)=T+274
T=-173 degree=100K(-173+273)---answer