A gas undergoes isothermal expansion at a constant temperature of 300K.If the gas absorbs 500J of heat during the process,calculate the work done by the gas

physics class 12 problems  physics class 12 numericals

Transcript

00:00Numerical number 1 of unit 16 first law of thermodynamics is a gas undergoes isothermal

00:14expansion at a constant temperature of 300 Kelvin if the gas absorbs 500 Joule of heat

00:23during the process calculate the work done by the gas so sab se peli hum data banayenge

00:33temperature is given as 300 Kelvin and heat is also given as 500 Joule

00:53so we have to find work done here process is isothermal so temperature is

01:05constant for calculation

01:11according to first law of thermodynamics delta Q is equal to delta U plus delta W is here

01:28is isothermal process so in isothermal process change in temperature is zero so delta U is

01:42also zero so delta Q will be equal to zero plus delta W so delta Q will be equal to work done

01:56means in this process heat supplied is equal to the work done so the value of heat supplied is 500 Joule

02:09so we will put 500 in place of delta Q or we can also write this is work done is equal to 500 Joule

02:25because in this process heat supplied is equal to the work done so this is our final answer