$A$ gas (Molar mass $= 280 \ g \ mol^{-1}$) was burnt in excess $O_2$ in a constant volume calorimeter and during combustion the temperature of calorimeter increased from $298.0 \ K$ to $298.45 \ K$. If the heat capacity of calorimeter is $2.5 \ kJ \ K^{-1}$ and enthalpy of combustion of gas is $9 \ kJ \ mol^{-1}$,then the amount of gas burnt is $...... \ g$. (Nearest Integer)
- A$35$
- B$34$
- C$33$
- D$32$
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The quantity of heat (in $J$) required to raise the temperature of $1.0 \, kg$ of ethanol from $293.45 \, K$ to the boiling point and then change the liquid to vapor at that temperature is closest to
[Given,boiling point of ethanol $351.45 \, K$. Specific heat capacity of liquid ethanol $2.44 \, J \, g^{-1} \, K^{-1}$. Latent heat of vaporisation of ethanol $855 \, J \, g^{-1}$ ]
[Given,boiling point of ethanol $351.45 \, K$. Specific heat capacity of liquid ethanol $2.44 \, J \, g^{-1} \, K^{-1}$. Latent heat of vaporisation of ethanol $855 \, J \, g^{-1}$ ]
MediumKVPY 2018
View SolutionFor the reaction $2X_{(g)} + Y_{(g)} \to 2Z_{(g)}$ at $298 \ K$,$\Delta U = -10.5 \ kJ$ and $\Delta S^o = -10.5 \ J/K$. Calculate $\Delta G^o$ for the reaction. Will the reaction be spontaneous or not? Explain.
Difficult
View SolutionFor a reaction,$X_{2(g)} + Y_{2(g)} \rightleftharpoons 2XY_{(g)}$. If $\Delta G_r^o = 20 \ kJ \ mol^{-1}$ and $\Delta S_r^o = -20 \ J \ K^{-1} \ mol^{-1}$ at $200 \ K$. Calculate $\Delta H_r^o$ at $400 \ K$ (Given $\Delta_r C_P = 20 \ J \ K^{-1} \ mol^{-1}$).
Medium
View SolutionMatch the following items in List-$I$ with their corresponding expressions in List-$II$.
List-$I$ List-$II$ $A$. At constant volume the change in internal energy of a system $I$. $W = -2.303 nRT \log \frac{V_f}{V_i}$ $B$. Isothermal irreversible change $II$. $W_{adiabatic} = \Delta U$ $C$. Isothermal reversible change $III$. $q_V = \Delta U$ $D$. Adiabatic change $IV$. $W = -p_{ex} (V_f - V_i)$ $V$. $\Delta U = \Delta H - \Delta nRT$
| List-$I$ | List-$II$ |
| $A$. At constant volume the change in internal energy of a system | $I$. $W = -2.303 nRT \log \frac{V_f}{V_i}$ |
| $B$. Isothermal irreversible change | $II$. $W_{adiabatic} = \Delta U$ |
| $C$. Isothermal reversible change | $III$. $q_V = \Delta U$ |
| $D$. Adiabatic change | $IV$. $W = -p_{ex} (V_f - V_i)$ |
| $V$. $\Delta U = \Delta H - \Delta nRT$ |
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