The work done by 3 moles of gas at 47°C to triple its volume at constant pressure is (\(R = 2\) cal mol\(^{-1}\) °C\(^{-1}\)):
At constant pressure, \(W = nR\Delta T\). Since volume triples, temperature also triples (from \(320\) K to \(960\) K), so \(\Delta T = 640\) K. Work \(W = 3 \times 2 \times 640 = 3840\) cal.
Consider statements (A) and (B) given below:
A. Thermodynamics deals with the process of conversion of heat into work only.
B. Heat given to a system and work done by the system are state variables in thermodynamics.
Choose the correct option.
Statement A is incorrect because thermodynamics also deals with other energy conversions. Statement B is incorrect because heat and work are path functions, not state variables.
Two moles of an ideal monoatomic gas undergoes an adiabatic process from temperature \(300\text{ K}\) to \(600\text{ K}\). Work done by this ideal gas in the process is
Work done in an adiabatic process is \(W = \frac{nR(T_1 - T_2)}{\gamma - 1}\). For monoatomic gas, \(\gamma = 5/3\). Substituting the parameters: \(W = \frac{2R(300 - 600)}{5/3 - 1} = \frac{-600R}{2/3} = -900R\).
110 J of heat is added to a gaseous system whose internal energy is increased by 40 J then amount of external work done is
According to the first law of thermodynamics, \(\Delta Q = \Delta U + W\). Substituting the values, \(110\text{ J} = 40\text{ J} + W \Rightarrow W = 70\text{ J}\).
Consider the following statements:
(A) Temperature of a body is related to its average internal energy, not to the kinetic energy of motion of its centre of mass.
(B) The first law of thermodynamics is based on law of conservation of energy applied to any system in which energy transfer from or to the surrounding (through heat and work) is taken into account.
Based on above information, pick correct option.
Statement (A) is correct because temperature measures internal, disordered molecular kinetic energy, not ordered bulk kinetic energy. Statement (B) is also correct because the first law of thermodynamics is simply the law of conservation of energy.
Assertion (A): It is possible for both the pressure and volume of a monoatomic ideal gas of a given amount to change simultaneously without causing the internal energy of the gas to change.
Reason (R): The internal energy of an ideal gas of a given amount remains constant if temperature does not change. It is possible to have a process in which pressure and volume are changed such that temperature remains constant.
For an ideal gas, internal energy \(U\) depends only on temperature \(T\). If \(U\) is constant, then \(T\) is constant. For a constant temperature process (isothermal), pressure \(P\) and volume \(V\) can change while \(T\) remains constant (as \(PV = nRT\)). Thus, both assertion and reason are true, and the reason correctly explains the assertion.
Assertion (A): Work done by a gas in isothermal expansion is more than the work done by the gas in the same expansion adiabatically.
Reason (R): Temperature remains constant in isothermal expansion but not in adiabatic expansion.
Assertion (A) is true. For the same volume expansion, work done \(W = int P dV\). In isothermal expansion, \(P\) drops slower than in adiabatic expansion (due to heat supply), so the area under the \(P-V\) curve is greater for isothermal. Reason (R) is also true and explains why \(P\) behaves differently, leading to different work done.
Assertion (A): During the melting of a slab of ice at \(273\text{ K}\) at \(1\text{ atm}\) positive work is done on the ice-water system by the atmosphere.
Reason (R): In above process, the internal energy of ice-water system increases.
When ice melts to water, its volume decreases \(\Delta V < 0\). Work done *by* the atmosphere *on* the system is \(-P\Delta V\), which is positive. So (A) is true. During melting, latent heat is absorbed, increasing internal energy \(\Delta U = Q - W\). Since \(Q\) is positive and \(W\) (work by system) is negative, \(\Delta U\) is positive. So (R) is true. However, the increase in internal energy is not the reason for the work done by the atmosphere; it's the volume change. So (R) does not explain (A).
Assertion (A): It is possible for both the pressure and volume of a monoatomic ideal gas of a given amount to change simultaneously without causing the internal energy of the gas to change.
Reason (R): The internal energy of an ideal gas of a given amount remains constant if temperature does not change. It is possible to have a process in which pressure and volume are changed such that temperature remains constant.
Concept: Internal energy of an ideal gas depends only on temperature.
Formula: For ideal gas, \( U = f(T) \). For monoatomic, \( U = \frac{3}{2} nRT \). Isothermal process implies \( T \) is constant.
Solution: If \( U \) is constant, then \( T \) is constant. An isothermal process allows simultaneous change in \( P \) and \( V \) while \( T \) (and thus \( U \)) remains constant. Reason correctly explains this.
Assertion (A): Work done by a gas in isothermal expansion is more than the work done by the gas in the same expansion adiabatically.
Reason (R): Temperature remains constant in isothermal expansion but not in adiabatic expansion.
Concept: Work done is area under P-V curve. Isothermal vs. Adiabatic expansion.
Formula: \( W = \int P dV \). Isothermal: \( PV = \text{constant} \). Adiabatic: \( PV^{\gamma} = \text{constant} \), where \( \gamma > 1 \).
Solution: During expansion from the same initial state to the same final volume, the pressure in isothermal process drops slower than in adiabatic process, leading to more work done. Temperature remains constant in isothermal and changes in adiabatic.