Question 371:
moderateWhen volume changes from $V$ to $2V$ at constant pressure $P$, then the change in internal energy will be: (1998)
Work done $\Delta W = P(2V - V) = PV$. Heat supplied $\Delta Q = n C_p \Delta T = \frac{\gamma PV}{\gamma-1}$. Thus, change in internal energy $\Delta U = \Delta Q - \Delta W = \frac{PV}{\gamma-1}$.
Question 372:
moderateA gas of volume changes $2\text{ litre}$ to $10\text{ litre}$ at constant temperature $300\text{ K}$, then the change in internal energy will be: (1998)
Since the process takes place at a constant temperature (isothermal), the internal energy of an ideal gas depends only on temperature, so the change in internal energy is zero.
Question 373:
moderateA sample of gas expands from volume $V_1$ to $V_2$. The amount of work done by the gas is greatest, when the expansion is: (1997)
On a $P-V$ diagram, the work done is represented by the area under the curve. For the same expansion volume, isobaric expansion maintains the highest pressure throughout, resulting in the maximum area and work done.
Question 374:
moderateAn ideal gas, undergoing adiabatic change, has which of the following pressure temperature relationship? (1996)
From the adiabatic relation $PV^\gamma = \text{constant}$ and the ideal gas law $PV = nRT$, eliminating volume yields $P^{1-\gamma} T^\gamma = \text{constant}$.
Question 375:
moderateA diatomic gas initially at $18^\circ\text{C}$ is compressed adiabatically to one eighth of its original volume. The temperature after compression will be: (1996)
Using the relation $T V^{\gamma-1} = \text{constant}$ with $\gamma = 1.4$ for a diatomic gas and $V_2 = V_1 / 8$, we find $T_2 = 291 \times (8)^{0.4} \approx 668.3\text{ K}$, which is $395.3^\circ\text{C}$.
Question 376:
moderateIn an adiabatic change, the pressure and temperature of a monoatomic gas are related as $P \propto T^C$ where $C$ equals: (1994)
From $P^{1-\gamma} T^\gamma = \text{constant}$, we get $P \propto T^{\frac{\gamma}{\gamma-1}}$. For a monoatomic gas, $\gamma = 5/3$, so $C = \frac{5/3}{5/3 - 1} = \frac{5}{2}$.
Question 377:
moderateA Carnot engine whose sink is at $300\text{ K}$ has an efficiency of $40\%$. By how much should the temperature of source be increased so as to increase its efficiency by $50\%$ of original efficiency? (2006)
Initial source temperature $T_1 = \frac{300}{1-0.4} = 500\text{ K}$. New efficiency $\eta' = 1.5 \times 0.4 = 0.6$. New source temperature $T_1' = \frac{300}{1-0.6} = 750\text{ K}$. Increase $\Delta T = 750 - 500 = 250\text{ K}$.
Question 378:
moderateAn ideal gas heat engine operates in Carnot cycle between $227\text{ }^\circ\text{C}$ and $127\text{ }^\circ\text{C}$. It absorbs $6 \times 10^4\text{ cal}$ of heat at higher temperature. Amount of heat converted to work is: (2005)
Temperatures are $T_1 = 500\text{ K}$ and $T_2 = 400\text{ K}$. Efficiency $\eta = 1 - \frac{400}{500} = 0.2$. Work $W = \eta Q_1 = 0.2 \times 6 \times 10^4 = 1.2 \times 10^4\text{ cal}$.
Question 379:
moderateAn ideal gas heat engine operates in a Carnot cycle. Between $227\text{ }^\circ\text{C}$ and $127\text{ }^\circ\text{C}$. It absorbs $6\text{ kcal}$ at the higher temperature. The amount of heat (in kcal) converted into work is equal to: (2003)
Efficiency $\eta = 1 - \frac{400}{500} = 0.2$. Work done $W = \eta Q_1 = 0.2 \times 6\text{ kcal} = 1.2\text{ kcal}$.
Question 380:
moderateThe efficiency of carnot engine is $50\%$ and temperature of sink is $500\text{ K}$. If temperature of source is kept constant and its efficiency raised to $60\%$, then the required temperature of the sink will be: (2002)
Source temperature $T_1 = \frac{500}{1-0.5} = 1000\text{ K}$. New sink temperature $T_2' = T_1(1-\eta_2) = 1000(1-0.6) = 400\text{ K}$.