Q.50. A certain gas is isothermally compressed to rd of its initial volume ( litre) by applying requ

Q: What is the correct answer for JEE Main Physics Q50?

The correct answer is 989. For an isothermal process, the bulk modulus $K = -V \frac{dP}{dV}$. For an ideal gas, $PV = nRT = \text{constant}$. So $P = \frac{C}{V}$. $dP = -\frac{C}{V^2} dV = -\frac{PV}{V^2} dV = -\frac{P}{V} dV$. So $K = -V \frac{(-P/V)dV}{dV} = P$. Thus, for an ideal gas undergoing an isothermal process, the bulk modulus is equal to the pressure. However, the question gives a constant value for the bulk mo

Q: Q.50. A certain gas is isothermally compressed to rd of its initial volume ( litre) by applying required pressure. If the bulk modulus of the gas is N/m, the magnitude of work done on the gas is…

The correct answer is 989. For an isothermal process, the bulk modulus $K = -V \frac{dP}{dV}$. For an ideal gas, $PV = nRT = \text{constant}$. So $P = \frac{C}{V}$. $dP = -\frac{C}{V^2} dV = -\frac{PV}{V^2} dV = -\frac{P}{V} dV$. So $K = -V \frac{(-P/V)dV}{dV} = P$. Thus, for an ideal gas undergoing an isothermal process, the bulk modulus is equal to the pressure. However, the question gives a constant value for the bulk mo

Question

Q.50. A certain gas is isothermally compressed to rd of its initial volume ( litre) by applying required pressure. If the bulk modulus of the gas is  N/m, the magnitude of work done on the gas is ________ J.

FirstInTest · Accepted Answer

For an isothermal process, the bulk modulus $K = -V \frac{dP}{dV}$. For an ideal gas, $PV = nRT = 	ext{constant}$. So $P = \frac{C}{V}$. $dP = -\frac{C}{V^2} dV = -\frac{PV}{V^2} dV = -\frac{P}{V} dV$. So $K = -V \frac{(-P/V)dV}{dV} = P$. Thus, for an ideal gas undergoing an isothermal process, the bulk modulus is equal to the pressure. However, the question gives a constant value for the bulk modulus. This implies the gas is not necessarily ideal or the bulk modulus is an average value. The work done during compression is $W = -\int_{V_1}^{V_2} P dV$. For an isothermal process, $PV = 	ext{constant}$. So $P = P_1 V_1 / V$. $W = -\int_{V_1}^{V_2} \frac{P_1 V_1}{V} dV = -P_1 V_1 \ln(\frac{V_2}{V_1})$. Given $V_1 = 3$ litre, $V_2 = V_1/3 = 1$ litre. The bulk modulus is given as $K = 3 	imes 10^5$ N/m$^2$. If we assume the bulk modulus is constant, then $K = P$. So $P = 3 	imes 10^5$ N/m$^2$. This is the pressure at which the compression occurs. If the bulk modulus is constant, then $P$ is constant, which means it's an isobaric process, not isothermal. This is a contradiction. The note in the solution states: "Bulk modulus of gas for isothermal process is not constant. In the given solution we have considered it as initial bulk modulus." This implies $P_1 = K = 3 	imes 10^5$ N/m$^2$. $V_1 = 3$ litre $= 3 	imes 10^{-3}$ m$^3$. $V_2 = 1$ litre $= 1 	imes 10^{-3}$ m$^3$. Work done $W = -P_1 V_1 \ln(\frac{V_2}{V_1}) = -(3 	imes 10^5 	ext{ N/m}^2) 	imes (3 	imes 10^{-3} 	ext{ m}^3) \ln(\frac{1 	imes 10^{-3}}{3 	imes 10^{-3}})$. $W = -(900 	ext{ J}) \ln(\frac{1}{3}) = -900 \ln(3^{-1}) = 900 \ln(3)$. $\ln(3) \approx 1.0986$. So $W = 900 	imes 1.0986 = 988.74$ J. The magnitude is 989 J (nearest integer). The assumption that initial pressure equals bulk modulus is crucial here, and it's a common simplification in such problems when a constant bulk modulus is given for an isothermal process.

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