Question 171:
moderateIf $Q$, $E$ and $W$ denote respectively the heat added, change in internal energy and the work done in a closed cycle process, then: (2004)
In a closed cycle process, the system returns to its initial state, meaning there is no net change in internal energy ($E = 0$ or $\Delta U = 0$).
Question 172:
moderateOne mole of an ideal gas at an initial temperatures of $T\text{ K}$ does $6\text{ R}$ joules of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $\frac{5}{3}$, the final temperatures of gas will be: (2004)
In an adiabatic process, $\Delta U = -W = -6R$. Also, $\Delta U = n C_v \Delta T = \frac{R}{\gamma - 1}(T_2 - T)$. With $\gamma = 5/3$, we get $\frac{3}{2}(T_2 - T) = -6$, leading to $T_2 = T - 4\text{ K}$.
Question 173:
moderateInitial pressure and volume of a gas are $P$ and $V$ respectively. First its volume is expanded to $4V$ by isothermal process and then again its volume makes to be $V$ by adiabatic process, then its final pressure is ($\gamma = 1.5$): (1999)
After isothermal expansion, pressure becomes $P/4$ at volume $4V$. Following adiabatic compression back to volume $V$, the final pressure is $P_3 = (P/4)(4)^{1.5} = P \times 4^{0.5} = 2P$.
Question 174:
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 175:
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 176:
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 177:
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 178:
moderateCarnot engine, having an efficiency of $\eta = 1/10$. As heat engine, is used as a refrigerator. If the work done on the system is $10\text{ J}$, the amount of energy absorbed from the reservoir at lower temperature is: (2015)
Using $\beta = \frac{1-\eta}{\eta} = 9$, the heat absorbed at lower temperature is $Q_2 = \beta W = 9 \times 10\text{ J} = 90\text{ J}$.
Question 179:
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: (2007)
Source temperature $T_1 = \frac{T_2}{1-\eta_1} = \frac{500}{0.5} = 1000\text{ K}$. New sink temperature $T_2' = T_1(1-\eta_2) = 1000(1 - 0.6) = 400\text{ K}$.
Question 180:
moderateAn engine has an efficiency of $1/6$. When the temperature of sink is reduced by $62\text{ }^\circ\text{C}$, its efficiency is doubled. Temperature of the source is (2007)
Initial efficiency $\eta_1 = 1 - \frac{T_2}{T_1} = \frac{1}{6}$. After reduction, $\eta_2 = 2(\frac{1}{6}) = \frac{1}{3} = 1 - \frac{T_2-62}{T_1}$. Solving these equations yields source temperature $T_1 = 372\text{ K} = 99\text{ }^\circ\text{C}$.