Tag: work done by an ideal gas in isothermal expansion

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Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

An ideal gas has initial volume V and pressure P. In doubling its volume the minimum work done will be in the following process(of given processes)

  1. Isobaric process

  2. Isothermal process

  3. Adiabatic process

  4. None of the above.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Work done is the area under the P-V curve. For a given expansion, the adiabatic curve is the steepest, resulting in the smallest area under the curve compared to isothermal or isobaric processes.

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

Two difference gases of molecular masses $M _1$ and $M _2$ are at the same temperature. What is the ratio of their mean square speeds?

  1. $\dfrac{M _1}{M _2}$

  2. $\dfrac{M _2}{M _1}$

  3. $\sqrt {\dfrac{M _1}{M _2}}$

  4. $\sqrt {\dfrac{M _2}{M _1}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mean squared speed $=\cfrac{3RT}{M}$

$\cfrac{V _1}{V _2}=\cfrac{M _1}{M _2}$

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

A diatomic gas which has initial volume of $10$ litre is isothermally compressed to $1/15^{th}$ of its original volume where initial pressure is $10^5$ Pascal. If temperature is $27^o$C then find the work done by gas.

  1. $-2.70\times 10^3$J

  2. $2.70\times 10^3$J

  3. $-1.35\times 10^3$J

  4. $1.35\times 10^3$J
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$w=nRT ln\left(\dfrac{v _2}{v _1}\right)$
$w=P _0V _0ln\left(\dfrac{v _2}{v _1}\right)$
$w=10^5\times 10\times 10^{-3}ln\left(\dfrac{1}{15}\right)$
$w=-2.70\times 10^3J$.

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

One mole of an ideal gas undergoes an isothermal change at temperature T so that its volume V is doubled. R is the molar gas constant. Work done by the gas during this change is :

  1. RT $\ln 4$

  2. RT $\ln 3$

  3. RT $ \ln 2$

  4. RT $ \ln 1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Under isothermal process work is given by the relation $W = RT \ln (\dfrac{V _{f}}{V _{i}})$
Therefore work will be $W = RT \ln(2)$

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

The slope of adiabatic curve is ________ than the slope of an isothermal curve.

  1. Greater.

  2. lesser

  3. data insufficient

  4. can be both a and b
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The slope of an adiabatic process is given by -gamma * P/V, while the slope of an isothermal process is -P/V. Since gamma (Cp/Cv) is always greater than 1 for an ideal gas, the adiabatic slope is steeper (greater in magnitude).

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

Three moles of an ideal gas $\left (C _{P} = \dfrac {7R}{2}\right )$ at pressure $P _{A}$ and temperature $T _{A}$ is isothermally expanded to twice the initial volume. The gas is then compressed at constant pressure to its original volume. Finally the gas is heated at constant volume to its original pressure $P _{A}$.
Calculate the net work done by the gas and the net heat supplied to the gas during the complete process.

  1. $0.579\ RT _{A}, \triangle Q = 0.579\ RT _{A}$.

  2. $79\ RT _{A}, \triangle Q = 0.679\ RT _{A}$.

  3. $0.9\ RT _{A}, \triangle Q = 0.779\ RT _{A}$.

  4. $0.7\ RT _{A}, \triangle Q = 0.979\ RT _{A}$.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The process involves isothermal expansion, isobaric compression, and isochoric heating. Calculating work for each step and summing them yields the net work; for a cycle, the heat supplied equals the work done if the internal energy change is zero, but here we sum the heat for each leg.

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

Two soap bubbles having radii $3\ cm$ and $4\ cm$ in vacuum, coalesce under isothermal conditions. The radius of the new bubble is

  1. $1\ cm$

  2. $5\ cm$

  3. $7\ cm$

  4. $3.5\ cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For soap bubbles coalescing isothermally, the total number of moles of air remains constant. Using PV = nRT, and P = 4T/R (where T is surface tension), the pressure-volume product is proportional to R^2. Since (4T/R1)*V1 + (4T/R2)*V2 = (4T/R_new)*V_new, and V is proportional to R^3, we get R1^2 + R2^2 = R_new^2.

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

One mole of an ideal gas u ndergoes a process:
$P = \dfrac{P _0}{1+(V _0/ V)^2}$.
Here $P _0$ and $V _0$ are constants. change in temperature of the gas when volume is changed from $V=V _0$ to $V = 2V _0$ is: 

  1. $-\dfrac{2P _0V _0}{5R}$

  2. $\dfrac{11P _0V _0}{10R}$

  3. $-\dfrac{5P _0V _0}{4R}$

  4. $P _0V _0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the ideal gas law PV = nRT, we substitute P = P0 / (1 + (V0/V)^2). Then T = PV/nR = (P0 * V) / (nR * (1 + (V0/V)^2)). Evaluating at V=V0 and V=2V0 allows calculating the change in temperature.

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

Let $Q$ and $W$ denote the amount of heat given to an ideal gas and the work done by it in an isothermal process.

  1. $Q = 0$

  2. $W = 0$

  3. $Q \neq W$

  4. $Q = W$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In an isothermal process for an ideal gas, the change in internal energy (dU) is zero because temperature is constant. According to the first law of thermodynamics (dQ = dU + dW), dQ must equal dW.

Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

Work done during isothermal expansion of one mole of an ideal gas $10$ atm to $1$ atm at $300\ K$ is

  1. $-4938.8\ J$

  2. $4938.8\ J$

  3. $-5744\ J$

  4. $6257.2\ J$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Work done in isothermal expansion is W = nRT ln(P1/P2). Using n=1, R=8.314 J/molK, T=300K, and ln(10/1) = 2.303, we get W = 1 * 8.314 * 300 * 2.303 = 5744 J. Since it is expansion, the gas does work, but the question asks for work done during expansion (often defined as positive by convention in physics).