Question 161:
moderate13. A monoatomic gas at a pressure $P$, having a volume $V$ expands isothermally to a volume $2V$ and then adiabatically to a volume $16V$. The final pressure of the gas is (take $\gamma = 5/3$): (2014)
For isothermal process $P_1V_1 = P_2V_2 \Rightarrow P \times V = P_2 \times 2V \Rightarrow P_2 = P/2$. For adiabatic process $P_2V_2^{\gamma} = P_3V_3^{\gamma} \Rightarrow (P/2)(2V)^{5/3} = P_3(16V)^{5/3} \Rightarrow P_3 = P/64$.
Question 162:
moderate14. During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its temperature. The ratio of $\frac{C_p}{C_v}$ for the gas is: (2013)
$P \propto T^3$ or $P T^{-3} = \text{constant}$. We know that for an adiabatic process $P^{1-\gamma} T^{\gamma} = \text{constant}$ or $P T^{\frac{\gamma}{1-\gamma}} = \text{constant}$. Comparing the powers of $T$, $\frac{\gamma}{1-\gamma} = -3 \Rightarrow \gamma = -3 + 3\gamma \Rightarrow 2\gamma = 3 \Rightarrow \gamma = 3/2$.
Question 163:
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 164:
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 165:
A mass of diatomic gas ($\gamma = 1.4$) at a pressure of $2\text{ atm}$ is compressed adiabatically so that its temperature rises from $27^\circ\text{C}$ to $927^\circ\text{C}$. The pressure of the gas in the final state is: (2011 Mains)
Using the adiabatic relation between pressure and temperature, $\frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^{\frac{\gamma}{\gamma-1}}$. Substituting $T_1 = 300\text{ K}$, $T_2 = 1200\text{ K}$, and $\gamma = 1.4$, we get $P_2 = 2 \times (4)^{3.5} = 256\text{ atm}$.
Question 166:
During an isothermal expansion, a confined ideal gas does $150\text{ J}$ of work against its surroundings. This implies that: (2011 Pre)
In an isothermal process, the temperature remains constant, so the change in internal energy is zero ($\Delta U = 0$). By the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W = 0 + 150\text{ J} = 150\text{ J}$, meaning heat is added.
Question 167:
A monoatomic gas at pressure $P_1$ and $V_1$ is compressed adiabatically to $\frac{1}{8}\text{th}$ its original volume. What is the final pressure of the gas? (2010 Mains)
For an adiabatic process, $P_1 V_1^\gamma = P_2 V_2^\gamma$. For a monoatomic gas, $\gamma = 5/3$. Substituting $V_2 = V_1 / 8$, we find $P_2 = P_1 (8)^{5/3} = 32 P_1$.
Question 168:
$\Delta U$ and $\Delta W$ represent the increase in internal energy and work done by the system respectively in a thermodynamical process, which of the following is true? (2010 Pre)
In an adiabatic process, no heat is exchanged with the surroundings ($\Delta Q = 0$). From the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$, which gives $\Delta U = -\Delta W$.
Question 169:
moderateIn thermodynamic processes which of the following statements is not true? (2009)
An isochoric process is one in which volume remains constant, not pressure. Therefore, statement (a) is incorrect.
Question 170:
moderateThe internal energy change in a system that has absorbed $2\text{ kcal}$ of heat and done $500\text{ J}$ of work is: (2009)
Using $\Delta Q = 2\text{ kcal} = 2000 \times 4.2\text{ J} = 8400\text{ J}$ and $\Delta W = 500\text{ J}$. From the first law, $\Delta U = \Delta Q - \Delta W = 8400 - 500 = 7900\text{ J}$.