Question 81:
easyStatement I: Internal energy of an ideal gas remains constant in an adiabatic process.
Statement II: In an adiabatic process, change in internal energy of a gas is equal to work done on or by the gas in the process.
In an adiabatic process, \(Q = 0\), so \(\Delta U = -W\), which means internal energy changes, so Statement I is incorrect. Statement II is correct since change in internal energy corresponds directly to the work done on or by the gas.
Question 82:
easyThe physical quantity that determines whether or not the given system A is in thermal equilibrium with another system B is called
Temperature is the physical quantity that determines thermal equilibrium. Two systems are in thermal equilibrium if and only if they are at the same temperature.
Question 83:
Equal masses of an ideal gas are sealed in two vessels one of pressure \(P_0\) and other of pressure \(2P_0\). If first vessel is at temperature of 400 K and the other is at 600 K. Find the ratio of volume of two container.
From the ideal gas law \(PV = nRT\), the volume is proportional to \(\frac{T}{P}\) for equal masses of the same gas. Thus, \(\frac{V_1}{V_2} = \frac{T_1}{T_2} \times \frac{P_2}{P_1} = \frac{400}{600} \times \frac{2P_0}{P_0} = \frac{4}{3}\).
Question 84:
moderateThe equation of state for 14 g nitrogen gas at a pressure P and temperature T, when occupying a volume V will be
Number of moles \(n = \frac{m}{M} = \frac{14\text{ g}}{28\text{ g/mol}} = 0.5\text{ mol}\). Substituting \(n = \frac{1}{2}\) in \(PV = nRT\) gives \(PV = \frac{1}{2}RT\).
Question 85:
easyFor $n$ mole of an ideal gas, the correct equation of $1^{\text{st}}$ law of thermodynamics corresponding to isobaric process will be (symbols have their usual meanings)
By first law, $Q = \Delta U + W$. In an isobaric process, the work done is $W = P\Delta V = nR\Delta T$. Therefore, both expressions are correct.
Question 86:
easyConsider the following thermodynamic parameters:
(a) Heat
(b) Internal energy
(c) Work
Which of the given parameters are path functions?
Heat and work depend on the path taken between states, whereas internal energy is a state function. Therefore, (a) and (c) are path functions.
Question 87:
easyIn ideal condition, the maximum efficiency that can be derived from a heat engine operating between $600\text{ K}$ reservoir and $200\text{ K}$ sink, is
Efficiency is given by $\eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}}$. Substituting the given values: $\eta = 1 -\frac{200}{600} =\frac{2}{3} \approx 66.67%$.
Question 88:
easyThe equation of state for \(14\text{ g}\) nitrogen gas at a pressure \(P\) and temperature \(T\), when occupying a volume \(V\) will be
The molecular mass of nitrogen gas \((\text{N}_2)\) is \(28\text{ g/mol}\). The number of moles is \(n = \frac{14}{28} = 0.5\). Thus, using \(PV = nRT\), we get \(PV = \frac{1}{2}RT\).
Question 89:
easyA monoatomic gas does 150 J of work in isothermal expansion. The heat supplied to the gas is
For an isothermal process, the change in internal energy is \( \Delta U = 0 \). According to the first law of thermodynamics, \( Q = \Delta U + W \), which gives \( Q = 0 + 150\text{ J} = 150\text{ J} \).
Question 90:
moderateIn thermodynamic processes, correct match of column-I with column-II is:
| Column-I (Type of process) | Column-II (Feature) |
| a. Isothermal | (iv) Temperature constant |
| b. Isobaric | (ii) Pressure constant |
| c. Isochoric | (i) Volume constant |
| d. Adiabatic | (iii) No heat flow between system and surroundings |
Isothermal process has constant temperature (a-iv). Isobaric has constant pressure (b-ii). Isochoric has constant volume (c-i). Adiabatic has no heat flow (d-iii). Matching these gives option D.