Question

Calculate magnitude of work done in calorie by one mole of an ideal gas subjected to the process as shown in figure. [Given: R = 2 Cal/mol / K, $\ln 2 = 0.7$]

• A. 640

Answer 1

Answer: (A)

640

Answer 2

The work done by an ideal gas is given by $W = \int P dV$. From the ideal gas law, $PV = nRT$, so $P = \frac{nRT}{V}$. The process is represented by a straight line in the T-V diagram, passing through points $(V_1, T_1) = (20 \text{ m}^3, 200 \text{ K})$ and $(V_2, T_2) = (40 \text{ m}^3, 800 \text{ K})$. The equation of the line is $T = mV + c$. The slope is $m = \frac{T_2 - T_1}{V_2 - V_1} = \frac{800 - 200}{40 - 20} = \frac{600}{20} = 30 \text{ K/m}^3$. Using point $(20, 200)$: $200 = 30(20) + c \implies 200 = 600 + c \implies c = -400 \text{ K}$. So, $T = 30V - 400$.

Now, substitute P in the work integral: $W = \int_{V_1}^{V_2} \frac{nRT}{V} dV$ $W = nR \int_{V_1}^{V_2} \frac{30V - 400}{V} dV$ $W = nR \int_{V_1}^{V_2} (30 - \frac{400}{V}) dV$ $W = nR \left[ 30V - 400 \ln|V| \right]_{V_1}^{V_2}$ $W = nR \left[ (30V_2 - 400 \ln V_2) - (30V_1 - 400 \ln V_1) \right]$ $W = nR \left[ 30(V_2 - V_1) - 400 (\ln V_2 - \ln V_1) \right]$ $W = nR \left[ 30(V_2 - V_1) - 400 \ln\left(\frac{V_2}{V_1}\right) \right]$

Given $n = 1$ mol, $R = 2$ Cal/mol/K, $V_1 = 20$ m³, $V_2 = 40$ m³, and $\ln 2 = 0.7$. $W = (1 \text{ mol}) \times (2 \text{ Cal/mol/K}) \left[ 30(40 \text{ m}^3 - 20 \text{ m}^3) - 400 \ln\left(\frac{40 \text{ m}^3}{20 \text{ m}^3}\right) \right]$ $W = 2 \text{ Cal/K} \left[ 30(20 \text{ m}^3) - 400 \ln(2) \right]$ $W = 2 \text{ Cal/K} \left[ 600 \text{ K} - 400 \times 0.7 \right]$ $W = 2 \text{ Cal/K} \left[ 600 \text{ K} - 280 \text{ K} \right]$ $W = 2 \text{ Cal/K} \left[ 320 \text{ K} \right]$ $W = 640 \text{ Cal}$.

Since the volume increases, the work done by the gas is positive. The magnitude of work done is therefore 640 Cal.