371555
In a given process \(dW = 0,\,\,dQ < 0\) then for a gas
1 Temperature increases
2 Volume decreases
3 Pressure decreases
4 Pressure increases
Explanation:
As \(d W=0\) so it is an isochoric process \(\dfrac{P}{T}=\) constant Given \(d q=d U < 0 \Rightarrow T\) decreases and hence \(P\) also decreases.
PHXI12:THERMODYNAMICS
371556
If \(R = \) universal gas constant, the amount of heat needed to rise the temperature of \(2\;\,mol\) of an ideal monoatomic gas from \(273\;K\) to \(373\;K\) when no work is done is
371558
Figure shows the pressure versus temperature curves for a given mass of a gas corresponding to two different volumes \({V_{1}}\) and \({V_{2}}\), then:
1 \({V_{1}>V_{2}}\)
2 \({V_{1} < V_{2}}\)
3 \({V_{1}=V_{2}}\)
4 The information is insufficient
Explanation:
Use \({P V=n R T}\) \(\frac{P}{T} = \frac{{nR}}{V}\) \(\therefore {\rm{ slope }} \propto \frac{1}{V}\) So correct Option is (1)
PHXI12:THERMODYNAMICS
371559
\(0.08\;kg\) air is heated at constant volume through \(5^\circ \,C\). The specific heat of air at constant volume is\(0.17\,kcal{\rm{/}}kg^\circ \,C\) and\(J = 4.18\,\,joule{\rm{/}}cal\). The change in its internal energy is approximately.
371555
In a given process \(dW = 0,\,\,dQ < 0\) then for a gas
1 Temperature increases
2 Volume decreases
3 Pressure decreases
4 Pressure increases
Explanation:
As \(d W=0\) so it is an isochoric process \(\dfrac{P}{T}=\) constant Given \(d q=d U < 0 \Rightarrow T\) decreases and hence \(P\) also decreases.
PHXI12:THERMODYNAMICS
371556
If \(R = \) universal gas constant, the amount of heat needed to rise the temperature of \(2\;\,mol\) of an ideal monoatomic gas from \(273\;K\) to \(373\;K\) when no work is done is
371558
Figure shows the pressure versus temperature curves for a given mass of a gas corresponding to two different volumes \({V_{1}}\) and \({V_{2}}\), then:
1 \({V_{1}>V_{2}}\)
2 \({V_{1} < V_{2}}\)
3 \({V_{1}=V_{2}}\)
4 The information is insufficient
Explanation:
Use \({P V=n R T}\) \(\frac{P}{T} = \frac{{nR}}{V}\) \(\therefore {\rm{ slope }} \propto \frac{1}{V}\) So correct Option is (1)
PHXI12:THERMODYNAMICS
371559
\(0.08\;kg\) air is heated at constant volume through \(5^\circ \,C\). The specific heat of air at constant volume is\(0.17\,kcal{\rm{/}}kg^\circ \,C\) and\(J = 4.18\,\,joule{\rm{/}}cal\). The change in its internal energy is approximately.
371555
In a given process \(dW = 0,\,\,dQ < 0\) then for a gas
1 Temperature increases
2 Volume decreases
3 Pressure decreases
4 Pressure increases
Explanation:
As \(d W=0\) so it is an isochoric process \(\dfrac{P}{T}=\) constant Given \(d q=d U < 0 \Rightarrow T\) decreases and hence \(P\) also decreases.
PHXI12:THERMODYNAMICS
371556
If \(R = \) universal gas constant, the amount of heat needed to rise the temperature of \(2\;\,mol\) of an ideal monoatomic gas from \(273\;K\) to \(373\;K\) when no work is done is
371558
Figure shows the pressure versus temperature curves for a given mass of a gas corresponding to two different volumes \({V_{1}}\) and \({V_{2}}\), then:
1 \({V_{1}>V_{2}}\)
2 \({V_{1} < V_{2}}\)
3 \({V_{1}=V_{2}}\)
4 The information is insufficient
Explanation:
Use \({P V=n R T}\) \(\frac{P}{T} = \frac{{nR}}{V}\) \(\therefore {\rm{ slope }} \propto \frac{1}{V}\) So correct Option is (1)
PHXI12:THERMODYNAMICS
371559
\(0.08\;kg\) air is heated at constant volume through \(5^\circ \,C\). The specific heat of air at constant volume is\(0.17\,kcal{\rm{/}}kg^\circ \,C\) and\(J = 4.18\,\,joule{\rm{/}}cal\). The change in its internal energy is approximately.
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI12:THERMODYNAMICS
371555
In a given process \(dW = 0,\,\,dQ < 0\) then for a gas
1 Temperature increases
2 Volume decreases
3 Pressure decreases
4 Pressure increases
Explanation:
As \(d W=0\) so it is an isochoric process \(\dfrac{P}{T}=\) constant Given \(d q=d U < 0 \Rightarrow T\) decreases and hence \(P\) also decreases.
PHXI12:THERMODYNAMICS
371556
If \(R = \) universal gas constant, the amount of heat needed to rise the temperature of \(2\;\,mol\) of an ideal monoatomic gas from \(273\;K\) to \(373\;K\) when no work is done is
371558
Figure shows the pressure versus temperature curves for a given mass of a gas corresponding to two different volumes \({V_{1}}\) and \({V_{2}}\), then:
1 \({V_{1}>V_{2}}\)
2 \({V_{1} < V_{2}}\)
3 \({V_{1}=V_{2}}\)
4 The information is insufficient
Explanation:
Use \({P V=n R T}\) \(\frac{P}{T} = \frac{{nR}}{V}\) \(\therefore {\rm{ slope }} \propto \frac{1}{V}\) So correct Option is (1)
PHXI12:THERMODYNAMICS
371559
\(0.08\;kg\) air is heated at constant volume through \(5^\circ \,C\). The specific heat of air at constant volume is\(0.17\,kcal{\rm{/}}kg^\circ \,C\) and\(J = 4.18\,\,joule{\rm{/}}cal\). The change in its internal energy is approximately.
371555
In a given process \(dW = 0,\,\,dQ < 0\) then for a gas
1 Temperature increases
2 Volume decreases
3 Pressure decreases
4 Pressure increases
Explanation:
As \(d W=0\) so it is an isochoric process \(\dfrac{P}{T}=\) constant Given \(d q=d U < 0 \Rightarrow T\) decreases and hence \(P\) also decreases.
PHXI12:THERMODYNAMICS
371556
If \(R = \) universal gas constant, the amount of heat needed to rise the temperature of \(2\;\,mol\) of an ideal monoatomic gas from \(273\;K\) to \(373\;K\) when no work is done is
371558
Figure shows the pressure versus temperature curves for a given mass of a gas corresponding to two different volumes \({V_{1}}\) and \({V_{2}}\), then:
1 \({V_{1}>V_{2}}\)
2 \({V_{1} < V_{2}}\)
3 \({V_{1}=V_{2}}\)
4 The information is insufficient
Explanation:
Use \({P V=n R T}\) \(\frac{P}{T} = \frac{{nR}}{V}\) \(\therefore {\rm{ slope }} \propto \frac{1}{V}\) So correct Option is (1)
PHXI12:THERMODYNAMICS
371559
\(0.08\;kg\) air is heated at constant volume through \(5^\circ \,C\). The specific heat of air at constant volume is\(0.17\,kcal{\rm{/}}kg^\circ \,C\) and\(J = 4.18\,\,joule{\rm{/}}cal\). The change in its internal energy is approximately.
