NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI12:THERMODYNAMICS
371534
The volume \((V)\) of a monatomic gas varies with its temperature \((T)\), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state \(A\) to state \(B\), is
1 \(\dfrac{1}{3}\)
2 \(\dfrac{2}{7}\)
3 \(\dfrac{2}{3}\)
4 \(\dfrac{2}{5}\)
Explanation:
The given process is isobaric. \(\begin{aligned}& W=P \Delta V=n R \Delta T \\& \Delta U=\dfrac{f}{2} n R \Delta T \\& \Delta Q=\Delta U+W=\left(\dfrac{f}{2}+1\right) n R \Delta T\end{aligned}\) Ratio \(\dfrac{W}{\Delta Q}=\dfrac{1}{\dfrac{f}{2}+1}=\dfrac{2}{f+2}=\dfrac{2}{5}\) For monoatomic gas \(f=3\)
NEET - 2018
PHXI12:THERMODYNAMICS
371535
In an isobaric process, the correct ratio is
371536
Two \(kg\) of water is converted into steam by boiling at atmospheric pressure. The volume changes from \(2 \times {10^{ - 3}}\;{m^3}\) to \(3.34\;{m^3}\). The work done by the system is about
371537
A gas at \({50 {~N} / {m}^{2}}\) pressure is compressed from \({10 m^{3}}\) to \({4 m^{3}}\) under constant pressure. Subsequently, it is given a heat of \(100\,J\). The internal energy of the gas will be
1 increased by \(100\,J\)
2 increased by \(200\,J\)
3 increased by \(400\,J\)
4 increased by \(300\,J\)
Explanation:
\({d Q=0}\) \(\therefore dU = - \Delta W = - P\left( {{V_2} - {V_1}} \right)\) \( = - 50(4 - 10) = 300\;\,J\) Additional heat given \({=100 {~J}}\) \({\therefore}\) Total increment in the internal energy \(d U=300+100=400 {~J}\)
371534
The volume \((V)\) of a monatomic gas varies with its temperature \((T)\), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state \(A\) to state \(B\), is
1 \(\dfrac{1}{3}\)
2 \(\dfrac{2}{7}\)
3 \(\dfrac{2}{3}\)
4 \(\dfrac{2}{5}\)
Explanation:
The given process is isobaric. \(\begin{aligned}& W=P \Delta V=n R \Delta T \\& \Delta U=\dfrac{f}{2} n R \Delta T \\& \Delta Q=\Delta U+W=\left(\dfrac{f}{2}+1\right) n R \Delta T\end{aligned}\) Ratio \(\dfrac{W}{\Delta Q}=\dfrac{1}{\dfrac{f}{2}+1}=\dfrac{2}{f+2}=\dfrac{2}{5}\) For monoatomic gas \(f=3\)
NEET - 2018
PHXI12:THERMODYNAMICS
371535
In an isobaric process, the correct ratio is
371536
Two \(kg\) of water is converted into steam by boiling at atmospheric pressure. The volume changes from \(2 \times {10^{ - 3}}\;{m^3}\) to \(3.34\;{m^3}\). The work done by the system is about
371537
A gas at \({50 {~N} / {m}^{2}}\) pressure is compressed from \({10 m^{3}}\) to \({4 m^{3}}\) under constant pressure. Subsequently, it is given a heat of \(100\,J\). The internal energy of the gas will be
1 increased by \(100\,J\)
2 increased by \(200\,J\)
3 increased by \(400\,J\)
4 increased by \(300\,J\)
Explanation:
\({d Q=0}\) \(\therefore dU = - \Delta W = - P\left( {{V_2} - {V_1}} \right)\) \( = - 50(4 - 10) = 300\;\,J\) Additional heat given \({=100 {~J}}\) \({\therefore}\) Total increment in the internal energy \(d U=300+100=400 {~J}\)
371534
The volume \((V)\) of a monatomic gas varies with its temperature \((T)\), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state \(A\) to state \(B\), is
1 \(\dfrac{1}{3}\)
2 \(\dfrac{2}{7}\)
3 \(\dfrac{2}{3}\)
4 \(\dfrac{2}{5}\)
Explanation:
The given process is isobaric. \(\begin{aligned}& W=P \Delta V=n R \Delta T \\& \Delta U=\dfrac{f}{2} n R \Delta T \\& \Delta Q=\Delta U+W=\left(\dfrac{f}{2}+1\right) n R \Delta T\end{aligned}\) Ratio \(\dfrac{W}{\Delta Q}=\dfrac{1}{\dfrac{f}{2}+1}=\dfrac{2}{f+2}=\dfrac{2}{5}\) For monoatomic gas \(f=3\)
NEET - 2018
PHXI12:THERMODYNAMICS
371535
In an isobaric process, the correct ratio is
371536
Two \(kg\) of water is converted into steam by boiling at atmospheric pressure. The volume changes from \(2 \times {10^{ - 3}}\;{m^3}\) to \(3.34\;{m^3}\). The work done by the system is about
371537
A gas at \({50 {~N} / {m}^{2}}\) pressure is compressed from \({10 m^{3}}\) to \({4 m^{3}}\) under constant pressure. Subsequently, it is given a heat of \(100\,J\). The internal energy of the gas will be
1 increased by \(100\,J\)
2 increased by \(200\,J\)
3 increased by \(400\,J\)
4 increased by \(300\,J\)
Explanation:
\({d Q=0}\) \(\therefore dU = - \Delta W = - P\left( {{V_2} - {V_1}} \right)\) \( = - 50(4 - 10) = 300\;\,J\) Additional heat given \({=100 {~J}}\) \({\therefore}\) Total increment in the internal energy \(d U=300+100=400 {~J}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXI12:THERMODYNAMICS
371534
The volume \((V)\) of a monatomic gas varies with its temperature \((T)\), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state \(A\) to state \(B\), is
1 \(\dfrac{1}{3}\)
2 \(\dfrac{2}{7}\)
3 \(\dfrac{2}{3}\)
4 \(\dfrac{2}{5}\)
Explanation:
The given process is isobaric. \(\begin{aligned}& W=P \Delta V=n R \Delta T \\& \Delta U=\dfrac{f}{2} n R \Delta T \\& \Delta Q=\Delta U+W=\left(\dfrac{f}{2}+1\right) n R \Delta T\end{aligned}\) Ratio \(\dfrac{W}{\Delta Q}=\dfrac{1}{\dfrac{f}{2}+1}=\dfrac{2}{f+2}=\dfrac{2}{5}\) For monoatomic gas \(f=3\)
NEET - 2018
PHXI12:THERMODYNAMICS
371535
In an isobaric process, the correct ratio is
371536
Two \(kg\) of water is converted into steam by boiling at atmospheric pressure. The volume changes from \(2 \times {10^{ - 3}}\;{m^3}\) to \(3.34\;{m^3}\). The work done by the system is about
371537
A gas at \({50 {~N} / {m}^{2}}\) pressure is compressed from \({10 m^{3}}\) to \({4 m^{3}}\) under constant pressure. Subsequently, it is given a heat of \(100\,J\). The internal energy of the gas will be
1 increased by \(100\,J\)
2 increased by \(200\,J\)
3 increased by \(400\,J\)
4 increased by \(300\,J\)
Explanation:
\({d Q=0}\) \(\therefore dU = - \Delta W = - P\left( {{V_2} - {V_1}} \right)\) \( = - 50(4 - 10) = 300\;\,J\) Additional heat given \({=100 {~J}}\) \({\therefore}\) Total increment in the internal energy \(d U=300+100=400 {~J}\)
