At N.T.P. one mole of diatomic gas is compressed adiabatically to half of its volume gamma = 1.41 .

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11.Thermodynamics

medium

At $N.T.P.$ one mole of diatomic gas is compressed adiabatically to half of its volume $\gamma = 1.41$. The work done on gas will be ....... $J$

$1280 $

$1610 $

$1815 $

$2025 $

Solution

(c)${T_2} = {T_1}{\left( {\frac{{{V_1}}}{{{V_2}}}} \right)^{\gamma – 1}} = 273{(2)^{0.41}}$$ = 273 \times 1.328 = 363K$
$W = \frac{{R({T_1} – {T_2})}}{{\gamma – 1}} = \frac{{8.31(273 – 363)}}{{1.41 – 1}}$$ = – \,1824$
==> $|W| \approx 1815 J$

Standard 11

Physics

The variation of pressure $P$ with volume $V$ for an ideal monatomic gas during an adiabatic process is shown in figure. At point $A$ the magnitude of rate of change of pressure with volume is

hard

In an adiabatic change, the pressure $P$ and temperature $T$ of a monoatomic gas are related by the relation $P \propto {T^C}$, where $C$ equals

medium

(AIIMS-2007)

Graph $A-B$ is an adiabatic curve. Choose the correct statement

medium

The initial pressure and volume of an ideal gas are $P_0$ and $V_0$. The final pressure of the gas when the gas is suddenly compressed to volume $\frac{ V _0}{4}$ will be (Given $\gamma=$ ratio of specific heats at constant pressure and at constant volume)

medium

(JEE MAIN-2023)

Match List $I$ with List $II$ :

List $I$ List $II$

$A$ Isothermal Process $I$ Work done by the gas decreases internal energy

$B$ Adiabatic Process $II$ No change in internal energy

$C$ Isochoric Process $III$ The heat absorbed goes partly to increase internal energy and partly to do work

$D$ Isobaric Process $IV$ No work is done on or by the gas

Choose the correct answer from the options given below :

medium

(JEE MAIN-2023)

List $I$	List $II$
$A$ Isothermal Process	$I$ Work done by the gas decreases internal energy
$B$ Adiabatic Process	$II$ No change in internal energy
$C$ Isochoric Process	$III$ The heat absorbed goes partly to increase internal energy and partly to do work
$D$ Isobaric Process	$IV$ No work is done on or by the gas

At $N.T.P.$ one mole of diatomic gas is compressed adiabatically to half of its volume $\gamma = 1.41$. The work done on gas will be ....... $J$

Solution

Similar Questions

The variation of pressure $P$ with volume $V$ for an ideal monatomic gas during an adiabatic process is shown in figure. At point $A$ the magnitude of rate of change of pressure with volume is

In an adiabatic change, the pressure $P$ and temperature $T$ of a monoatomic gas are related by the relation $P \propto {T^C}$, where $C$ equals

Graph $A-B$ is an adiabatic curve. Choose the correct statement

The initial pressure and volume of an ideal gas are $P_0$ and $V_0$. The final pressure of the gas when the gas is suddenly compressed to volume $\frac{ V _0}{4}$ will be (Given $\gamma=$ ratio of specific heats at constant pressure and at constant volume)

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