NEET Test Series from KOTA - 10 Papers In MS WORD
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Thermodynamics
272411
If an endothermic reaction occurs spontaneously at constant temperature $T$ and $P$, then which of the following is true?
1 $\Delta \mathrm{G}>0$
2 $\Delta \mathrm{H}<0$
3 $\Delta \mathrm{S}>0$
4 $\Delta \mathrm{S}<0$
Explanation:
For a reaction to take place spontaneously the value of $\Delta \mathrm{G}$ must be negative i.e. $\Delta \mathrm{G}<0$. Now, $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$. - As the reaction is endothermic, so value of $\Delta \mathrm{H}$ must be positive, i.e. $\Delta \mathrm{H}>0$. Hence, to have a negative $\Delta \mathrm{G}$. $\Delta \mathrm{H}<\mathrm{T} \Delta \mathrm{S} . \quad$ As $\mathrm{T} \\mathrm{P}$ are constant. - $T \Delta S$ must be positive to give the total value a negative $\operatorname{sign}$. Hence, $\Delta \mathrm{S}>0$.
Gibb's Helmholtz Equation- $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ Differentiate this equation w.r.t. temperature at constant pressure, $\left(\frac{\partial \Delta \mathrm{G}}{\partial \mathrm{T}}\right)_{\mathrm{p}}=\left(\frac{\partial \mathrm{G}_{\mathrm{y}}}{\partial \mathrm{T}}\right)_{\mathrm{p}}-\left(\frac{\partial \mathrm{G}_x}{\partial \mathrm{T}}\right)_p$ $=-\mathrm{S}_{\mathrm{y}}-\left(-\mathrm{S}_{\mathrm{x}}\right)$ $=-\left(\mathrm{S}_{\mathrm{y}}-\mathrm{S}_{\mathrm{x}}\right)=-\Delta \mathrm{S}$ Where, $\Delta S$ change in entropy On combining equation (1) \& (3), we get- $\Delta \mathrm{G}=\Delta \mathrm{H}+\mathrm{T}\left(\frac{\partial(\Delta \mathrm{G})}{\partial \mathrm{T}}\right)_{\mathrm{p}}$ Equation (4) is an alternative form of Gibbs Helmholtz equation, Dividing equation (4) by $\mathrm{T}^2$, we get $\frac{\Delta \mathrm{G}}{\mathrm{T}^2}=\frac{\Delta \mathrm{H}}{\mathrm{T}^2}+\frac{1}{\mathrm{~T}}\left[\frac{\partial(\Delta \mathrm{G})}{\partial \mathrm{T}}\right]_{\mathrm{p}}$ On rearrangement, we get $\left[\frac{\partial(\Delta \mathrm{G})}{\partial \mathrm{T}}\right]_{\mathrm{p}}=-\frac{\Delta \mathrm{H}}{\mathrm{T}^2}$ $\therefore \Delta \mathrm{H}=-\mathrm{T}^2\left[\frac{\partial(\Delta \mathrm{G})}{\partial \mathrm{T}}\right]_{\mathrm{P}}$
VITEEE- 2007
Thermodynamics
272416
Which of the following conditions lead to a spontaneous process?
1 $\Delta \mathrm{H}$ is + ve ; $\mathrm{T} \Delta \mathrm{S}$ is - ve
3 Both $\Delta \mathrm{H}$ and $\mathrm{T} \Delta \mathrm{S}$ are + ve but $\Delta \mathrm{H}>\mathrm{T} \Delta \mathrm{S}$
4 Both $\Delta \mathrm{H}$ and $\mathrm{T} \Delta \mathrm{S}$ are + ve but $\mathrm{T} \Delta \mathrm{S}>\Delta \mathrm{H}$
Explanation:
We know that, $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ Where, $\Delta \mathrm{G}=\mathrm{Gibb}$ 's free energy change $\Delta \mathrm{H}=$ enthalpy change $\Delta \mathrm{S}=$ entropy change. For the spontaneous process, $\Delta \mathrm{G}$ should be negative which is possible in only option (d) because both $\Delta \mathrm{H}$ and $\mathrm{T} \Delta \mathrm{S}$ are positive and $\mathrm{T} \Delta \mathrm{S}>\Delta \mathrm{H}$ due to which $\Delta \mathrm{G}<0$.
4 $\frac{\mathrm{P}}{\mathrm{T}}=\frac{4}{500}=0.0080$ $\because \mathrm{P} / \mathrm{T}$ ratio is highest at $4 \mathrm{~atm}$ and $500 \mathrm{~K}$. $\therefore$ Density is maximum at this condition.
Explanation:
For ideal gas equation, $P=\mathrm{dRT}$ (d = Density) $\therefore \quad d \propto \frac{P}{T}$ (a.) $\frac{\mathrm{P}}{\mathrm{T}}=\frac{2}{600}=0.0033$ (b.) $\frac{\mathrm{P}}{\mathrm{T}}=\frac{0.5}{273}=0.00183$ (c.) $\frac{\mathrm{P}}{\mathrm{T}}=\frac{6}{1092}=0.00549$ (d.) $\frac{\mathrm{P}}{\mathrm{T}}=\frac{4}{500}=0.0080$ $\because \mathrm{P} / \mathrm{T}$ ratio is highest at $4 \mathrm{~atm}$ and $500 \mathrm{~K}$. $\therefore$ Density is maximum at this condition.
272411
If an endothermic reaction occurs spontaneously at constant temperature $T$ and $P$, then which of the following is true?
1 $\Delta \mathrm{G}>0$
2 $\Delta \mathrm{H}<0$
3 $\Delta \mathrm{S}>0$
4 $\Delta \mathrm{S}<0$
Explanation:
For a reaction to take place spontaneously the value of $\Delta \mathrm{G}$ must be negative i.e. $\Delta \mathrm{G}<0$. Now, $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$. - As the reaction is endothermic, so value of $\Delta \mathrm{H}$ must be positive, i.e. $\Delta \mathrm{H}>0$. Hence, to have a negative $\Delta \mathrm{G}$. $\Delta \mathrm{H}<\mathrm{T} \Delta \mathrm{S} . \quad$ As $\mathrm{T} \\mathrm{P}$ are constant. - $T \Delta S$ must be positive to give the total value a negative $\operatorname{sign}$. Hence, $\Delta \mathrm{S}>0$.
Gibb's Helmholtz Equation- $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ Differentiate this equation w.r.t. temperature at constant pressure, $\left(\frac{\partial \Delta \mathrm{G}}{\partial \mathrm{T}}\right)_{\mathrm{p}}=\left(\frac{\partial \mathrm{G}_{\mathrm{y}}}{\partial \mathrm{T}}\right)_{\mathrm{p}}-\left(\frac{\partial \mathrm{G}_x}{\partial \mathrm{T}}\right)_p$ $=-\mathrm{S}_{\mathrm{y}}-\left(-\mathrm{S}_{\mathrm{x}}\right)$ $=-\left(\mathrm{S}_{\mathrm{y}}-\mathrm{S}_{\mathrm{x}}\right)=-\Delta \mathrm{S}$ Where, $\Delta S$ change in entropy On combining equation (1) \& (3), we get- $\Delta \mathrm{G}=\Delta \mathrm{H}+\mathrm{T}\left(\frac{\partial(\Delta \mathrm{G})}{\partial \mathrm{T}}\right)_{\mathrm{p}}$ Equation (4) is an alternative form of Gibbs Helmholtz equation, Dividing equation (4) by $\mathrm{T}^2$, we get $\frac{\Delta \mathrm{G}}{\mathrm{T}^2}=\frac{\Delta \mathrm{H}}{\mathrm{T}^2}+\frac{1}{\mathrm{~T}}\left[\frac{\partial(\Delta \mathrm{G})}{\partial \mathrm{T}}\right]_{\mathrm{p}}$ On rearrangement, we get $\left[\frac{\partial(\Delta \mathrm{G})}{\partial \mathrm{T}}\right]_{\mathrm{p}}=-\frac{\Delta \mathrm{H}}{\mathrm{T}^2}$ $\therefore \Delta \mathrm{H}=-\mathrm{T}^2\left[\frac{\partial(\Delta \mathrm{G})}{\partial \mathrm{T}}\right]_{\mathrm{P}}$
VITEEE- 2007
Thermodynamics
272416
Which of the following conditions lead to a spontaneous process?
1 $\Delta \mathrm{H}$ is + ve ; $\mathrm{T} \Delta \mathrm{S}$ is - ve
3 Both $\Delta \mathrm{H}$ and $\mathrm{T} \Delta \mathrm{S}$ are + ve but $\Delta \mathrm{H}>\mathrm{T} \Delta \mathrm{S}$
4 Both $\Delta \mathrm{H}$ and $\mathrm{T} \Delta \mathrm{S}$ are + ve but $\mathrm{T} \Delta \mathrm{S}>\Delta \mathrm{H}$
Explanation:
We know that, $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ Where, $\Delta \mathrm{G}=\mathrm{Gibb}$ 's free energy change $\Delta \mathrm{H}=$ enthalpy change $\Delta \mathrm{S}=$ entropy change. For the spontaneous process, $\Delta \mathrm{G}$ should be negative which is possible in only option (d) because both $\Delta \mathrm{H}$ and $\mathrm{T} \Delta \mathrm{S}$ are positive and $\mathrm{T} \Delta \mathrm{S}>\Delta \mathrm{H}$ due to which $\Delta \mathrm{G}<0$.
4 $\frac{\mathrm{P}}{\mathrm{T}}=\frac{4}{500}=0.0080$ $\because \mathrm{P} / \mathrm{T}$ ratio is highest at $4 \mathrm{~atm}$ and $500 \mathrm{~K}$. $\therefore$ Density is maximum at this condition.
Explanation:
For ideal gas equation, $P=\mathrm{dRT}$ (d = Density) $\therefore \quad d \propto \frac{P}{T}$ (a.) $\frac{\mathrm{P}}{\mathrm{T}}=\frac{2}{600}=0.0033$ (b.) $\frac{\mathrm{P}}{\mathrm{T}}=\frac{0.5}{273}=0.00183$ (c.) $\frac{\mathrm{P}}{\mathrm{T}}=\frac{6}{1092}=0.00549$ (d.) $\frac{\mathrm{P}}{\mathrm{T}}=\frac{4}{500}=0.0080$ $\because \mathrm{P} / \mathrm{T}$ ratio is highest at $4 \mathrm{~atm}$ and $500 \mathrm{~K}$. $\therefore$ Density is maximum at this condition.
272411
If an endothermic reaction occurs spontaneously at constant temperature $T$ and $P$, then which of the following is true?
1 $\Delta \mathrm{G}>0$
2 $\Delta \mathrm{H}<0$
3 $\Delta \mathrm{S}>0$
4 $\Delta \mathrm{S}<0$
Explanation:
For a reaction to take place spontaneously the value of $\Delta \mathrm{G}$ must be negative i.e. $\Delta \mathrm{G}<0$. Now, $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$. - As the reaction is endothermic, so value of $\Delta \mathrm{H}$ must be positive, i.e. $\Delta \mathrm{H}>0$. Hence, to have a negative $\Delta \mathrm{G}$. $\Delta \mathrm{H}<\mathrm{T} \Delta \mathrm{S} . \quad$ As $\mathrm{T} \\mathrm{P}$ are constant. - $T \Delta S$ must be positive to give the total value a negative $\operatorname{sign}$. Hence, $\Delta \mathrm{S}>0$.
Gibb's Helmholtz Equation- $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ Differentiate this equation w.r.t. temperature at constant pressure, $\left(\frac{\partial \Delta \mathrm{G}}{\partial \mathrm{T}}\right)_{\mathrm{p}}=\left(\frac{\partial \mathrm{G}_{\mathrm{y}}}{\partial \mathrm{T}}\right)_{\mathrm{p}}-\left(\frac{\partial \mathrm{G}_x}{\partial \mathrm{T}}\right)_p$ $=-\mathrm{S}_{\mathrm{y}}-\left(-\mathrm{S}_{\mathrm{x}}\right)$ $=-\left(\mathrm{S}_{\mathrm{y}}-\mathrm{S}_{\mathrm{x}}\right)=-\Delta \mathrm{S}$ Where, $\Delta S$ change in entropy On combining equation (1) \& (3), we get- $\Delta \mathrm{G}=\Delta \mathrm{H}+\mathrm{T}\left(\frac{\partial(\Delta \mathrm{G})}{\partial \mathrm{T}}\right)_{\mathrm{p}}$ Equation (4) is an alternative form of Gibbs Helmholtz equation, Dividing equation (4) by $\mathrm{T}^2$, we get $\frac{\Delta \mathrm{G}}{\mathrm{T}^2}=\frac{\Delta \mathrm{H}}{\mathrm{T}^2}+\frac{1}{\mathrm{~T}}\left[\frac{\partial(\Delta \mathrm{G})}{\partial \mathrm{T}}\right]_{\mathrm{p}}$ On rearrangement, we get $\left[\frac{\partial(\Delta \mathrm{G})}{\partial \mathrm{T}}\right]_{\mathrm{p}}=-\frac{\Delta \mathrm{H}}{\mathrm{T}^2}$ $\therefore \Delta \mathrm{H}=-\mathrm{T}^2\left[\frac{\partial(\Delta \mathrm{G})}{\partial \mathrm{T}}\right]_{\mathrm{P}}$
VITEEE- 2007
Thermodynamics
272416
Which of the following conditions lead to a spontaneous process?
1 $\Delta \mathrm{H}$ is + ve ; $\mathrm{T} \Delta \mathrm{S}$ is - ve
3 Both $\Delta \mathrm{H}$ and $\mathrm{T} \Delta \mathrm{S}$ are + ve but $\Delta \mathrm{H}>\mathrm{T} \Delta \mathrm{S}$
4 Both $\Delta \mathrm{H}$ and $\mathrm{T} \Delta \mathrm{S}$ are + ve but $\mathrm{T} \Delta \mathrm{S}>\Delta \mathrm{H}$
Explanation:
We know that, $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ Where, $\Delta \mathrm{G}=\mathrm{Gibb}$ 's free energy change $\Delta \mathrm{H}=$ enthalpy change $\Delta \mathrm{S}=$ entropy change. For the spontaneous process, $\Delta \mathrm{G}$ should be negative which is possible in only option (d) because both $\Delta \mathrm{H}$ and $\mathrm{T} \Delta \mathrm{S}$ are positive and $\mathrm{T} \Delta \mathrm{S}>\Delta \mathrm{H}$ due to which $\Delta \mathrm{G}<0$.
4 $\frac{\mathrm{P}}{\mathrm{T}}=\frac{4}{500}=0.0080$ $\because \mathrm{P} / \mathrm{T}$ ratio is highest at $4 \mathrm{~atm}$ and $500 \mathrm{~K}$. $\therefore$ Density is maximum at this condition.
Explanation:
For ideal gas equation, $P=\mathrm{dRT}$ (d = Density) $\therefore \quad d \propto \frac{P}{T}$ (a.) $\frac{\mathrm{P}}{\mathrm{T}}=\frac{2}{600}=0.0033$ (b.) $\frac{\mathrm{P}}{\mathrm{T}}=\frac{0.5}{273}=0.00183$ (c.) $\frac{\mathrm{P}}{\mathrm{T}}=\frac{6}{1092}=0.00549$ (d.) $\frac{\mathrm{P}}{\mathrm{T}}=\frac{4}{500}=0.0080$ $\because \mathrm{P} / \mathrm{T}$ ratio is highest at $4 \mathrm{~atm}$ and $500 \mathrm{~K}$. $\therefore$ Density is maximum at this condition.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Thermodynamics
272411
If an endothermic reaction occurs spontaneously at constant temperature $T$ and $P$, then which of the following is true?
1 $\Delta \mathrm{G}>0$
2 $\Delta \mathrm{H}<0$
3 $\Delta \mathrm{S}>0$
4 $\Delta \mathrm{S}<0$
Explanation:
For a reaction to take place spontaneously the value of $\Delta \mathrm{G}$ must be negative i.e. $\Delta \mathrm{G}<0$. Now, $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$. - As the reaction is endothermic, so value of $\Delta \mathrm{H}$ must be positive, i.e. $\Delta \mathrm{H}>0$. Hence, to have a negative $\Delta \mathrm{G}$. $\Delta \mathrm{H}<\mathrm{T} \Delta \mathrm{S} . \quad$ As $\mathrm{T} \\mathrm{P}$ are constant. - $T \Delta S$ must be positive to give the total value a negative $\operatorname{sign}$. Hence, $\Delta \mathrm{S}>0$.
Gibb's Helmholtz Equation- $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ Differentiate this equation w.r.t. temperature at constant pressure, $\left(\frac{\partial \Delta \mathrm{G}}{\partial \mathrm{T}}\right)_{\mathrm{p}}=\left(\frac{\partial \mathrm{G}_{\mathrm{y}}}{\partial \mathrm{T}}\right)_{\mathrm{p}}-\left(\frac{\partial \mathrm{G}_x}{\partial \mathrm{T}}\right)_p$ $=-\mathrm{S}_{\mathrm{y}}-\left(-\mathrm{S}_{\mathrm{x}}\right)$ $=-\left(\mathrm{S}_{\mathrm{y}}-\mathrm{S}_{\mathrm{x}}\right)=-\Delta \mathrm{S}$ Where, $\Delta S$ change in entropy On combining equation (1) \& (3), we get- $\Delta \mathrm{G}=\Delta \mathrm{H}+\mathrm{T}\left(\frac{\partial(\Delta \mathrm{G})}{\partial \mathrm{T}}\right)_{\mathrm{p}}$ Equation (4) is an alternative form of Gibbs Helmholtz equation, Dividing equation (4) by $\mathrm{T}^2$, we get $\frac{\Delta \mathrm{G}}{\mathrm{T}^2}=\frac{\Delta \mathrm{H}}{\mathrm{T}^2}+\frac{1}{\mathrm{~T}}\left[\frac{\partial(\Delta \mathrm{G})}{\partial \mathrm{T}}\right]_{\mathrm{p}}$ On rearrangement, we get $\left[\frac{\partial(\Delta \mathrm{G})}{\partial \mathrm{T}}\right]_{\mathrm{p}}=-\frac{\Delta \mathrm{H}}{\mathrm{T}^2}$ $\therefore \Delta \mathrm{H}=-\mathrm{T}^2\left[\frac{\partial(\Delta \mathrm{G})}{\partial \mathrm{T}}\right]_{\mathrm{P}}$
VITEEE- 2007
Thermodynamics
272416
Which of the following conditions lead to a spontaneous process?
1 $\Delta \mathrm{H}$ is + ve ; $\mathrm{T} \Delta \mathrm{S}$ is - ve
3 Both $\Delta \mathrm{H}$ and $\mathrm{T} \Delta \mathrm{S}$ are + ve but $\Delta \mathrm{H}>\mathrm{T} \Delta \mathrm{S}$
4 Both $\Delta \mathrm{H}$ and $\mathrm{T} \Delta \mathrm{S}$ are + ve but $\mathrm{T} \Delta \mathrm{S}>\Delta \mathrm{H}$
Explanation:
We know that, $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ Where, $\Delta \mathrm{G}=\mathrm{Gibb}$ 's free energy change $\Delta \mathrm{H}=$ enthalpy change $\Delta \mathrm{S}=$ entropy change. For the spontaneous process, $\Delta \mathrm{G}$ should be negative which is possible in only option (d) because both $\Delta \mathrm{H}$ and $\mathrm{T} \Delta \mathrm{S}$ are positive and $\mathrm{T} \Delta \mathrm{S}>\Delta \mathrm{H}$ due to which $\Delta \mathrm{G}<0$.
4 $\frac{\mathrm{P}}{\mathrm{T}}=\frac{4}{500}=0.0080$ $\because \mathrm{P} / \mathrm{T}$ ratio is highest at $4 \mathrm{~atm}$ and $500 \mathrm{~K}$. $\therefore$ Density is maximum at this condition.
Explanation:
For ideal gas equation, $P=\mathrm{dRT}$ (d = Density) $\therefore \quad d \propto \frac{P}{T}$ (a.) $\frac{\mathrm{P}}{\mathrm{T}}=\frac{2}{600}=0.0033$ (b.) $\frac{\mathrm{P}}{\mathrm{T}}=\frac{0.5}{273}=0.00183$ (c.) $\frac{\mathrm{P}}{\mathrm{T}}=\frac{6}{1092}=0.00549$ (d.) $\frac{\mathrm{P}}{\mathrm{T}}=\frac{4}{500}=0.0080$ $\because \mathrm{P} / \mathrm{T}$ ratio is highest at $4 \mathrm{~atm}$ and $500 \mathrm{~K}$. $\therefore$ Density is maximum at this condition.