- Home
- Standard 11
- Physics
11.Thermodynamics
medium
A gas is compressed from a volume of $2\,m^3$ to a volume of $1\, m^3$ at a constant pressure of $100\, N/m^2$. Then it is heated at constant volume by supplying $150\, J$ of energy. As a result, the internal energy of the gas
A
increases by $250\, J$
B
decreases by $250\, J$
C
increases by $50\, J$
D
decreases by $50\, J$
(JEE MAIN-2014)
Solution
As we know, $\Delta Q=\Delta u+\Delta w$
(Ist law of thermodynamics) $\Rightarrow \Delta \mathrm{Q}=\Delta \mathrm{u}+\mathrm{P} \Delta \mathrm{v}$
or $150=\Delta u+100(1-2)$
$=\Delta u-100$
$\therefore \Delta u=150+100=250 \mathrm{J}$
Thus the internal energy of the gas increases by $250 \mathrm{J}$
Standard 11
Physics
Similar Questions
In Column $-I$ processes and in Column $-II$ formulas of work are given. Match them appropriately :
| Column $-I$ | Column $-II$ |
| $(a)$ Isothermal process | $(i)$ $W = \frac{{\mu R({T_1} – {T_2})}}{{\gamma – 1}}$ |
| $(b)$ Adiabatic process | $(ii)$ $W = P\Delta V$ |
| $(iii)$ $W = 2.303\,\mu RT\log \left( {\frac{{{V_2}}}{{{V_1}}}} \right)$ |
easy
