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Physics Question 34 – JEE-MAIN 2026

An ideal gas at pressure P and temperature T is expanding such that PT3=constant. The coefficient of volume expansion of the gas is _______.

The coefficient of volume expansion β for a process is defined as the fractional change in volume per unit change in temperature, i.e., β=1VdVdT.

Step 1: Relate Volume, Pressure, and Temperature✦ Active

The ideal gas law is PV=nRT. From this, we can express volume V as V=nRTP. The given process condition is PT3=K, where K is a constant. We can express pressure P as P=KT3. Substituting this into the ideal gas law gives the relationship between V and T for this process:

V=nRT(K/T3)=nRT4K
Step 2: Calculate the Derivative of Volume with respect to Temperature○ Expand

The coefficient of volume expansion β is defined as β=1V(dVdT). First, differentiate the expression for V with respect to T:

dVdT=ddT(nRT4K)=nRK(4T3)
💡 Teacher's Secret Hint

Remember to treat n, R, and K as constants during differentiation.

Step 3: Determine the Coefficient of Volume Expansion○ Expand

Now, substitute the expressions for V and dVdT into the definition of β:

β=1V(dVdT)=1(nRT4K)(4nRT3K) β=KnRT44nRT3K=4T

Thus, the coefficient of volume expansion of the gas is 4T.

💡 Teacher's Secret Hint

Alternatively, one could use logarithmic differentiation: lnV=ln(nR)+lnTlnP and lnP+3lnT=lnK. Differentiating both with respect to T and combining them yields the same result.

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