Thermodynamics is a scoring chapter in JEE physics. Here are some Some helpful Tips for Thermodynamics to solve the problem correctly and quickly

**Tip 1**. Always make sure you Heat(Q) ,Internal Energy(U) & Work(W) with correct sign in equation

$\Delta U=Q-W$

W is positive when work done by the system(Increase in Volume)

W is negative when work done on the system(Decrease in Volume)

Q is positive when heat is given to the system

Q is negative when heat is taken from the system

**Tip 2**. Work done in a cycle is equal to the area enclosed on the PV diagram

**Tip 3**. Change in Internal energy is zero for a cycle

$\Delta U=0$

So Net Heat=Net Work done

$Q=W$

**Tip 4**.All Bodies emit radiation at all temperature.Radiation emitted are of different frequencies.Frequency at which most radiation are emitted are directly proportional to Temperature of the body.

**Tip 5**. **Isothermal Process**

PV=constant

$VdP+PdV=0$

$-\frac {VdP}{dV}=P$

Now Bulk Modulus= $-\frac {VdP}{dV}$

Therefore B_{i}=P

Work done is given by

$W=nRT ln \frac {V_2}{V_1}$

or

$W=nRT ln \frac {P_1}{P_2}$

$\Delta U=0$

$Q =W$

**Adiabatic Process**

$PV^{\lambda}=constant$

$V^{\lambda} (dP) + P \lambda V^{\lambda -1}=0$

So $B_a=\lambda P$

Work done is given by

$W=\frac {P_2V_2 – P_1V_1}{1-\lambda }$

Or

$W=nr\frac {T_1 – T_2}{1-\lambda }$

$\Delta Q=0$

$\Delta U =-W$

**Isochoric process**

$\Delta V=0$ , so $\Delta W=0$

$\Delta U =Q$

$\frac {P}{T} = constant$

**Isobaric Process**

$W= P(V_2 -V_1$

$\frac {V}{T} = constant$

$\Delta U=Q-W$

**Tip 6**. if A sold sphere and a hollow sphere having same material and size are heated to same temperature and allowed to cool in the same surrounding ,it has been found that hollow sphere cools faster than solid sphere

**Tip 7**

$C_P- C_V=R$

$\Delta = nC_V \Delta T$